The Truth Table Method & The Truth Tree Method

1 Introduction

1. Unlike the Big 8 Method and like the Method of Derivation, the Truth Table method and the Truth Tree method can be applied to inferences of any length; each will work no matter how many (or how few) premises an inference has.

2. The method of Derivation tells us that, if we find a derivation, the conclusion follows validly from the premises, but it remains silent when we cannot find a derivation; perhaps we have failed to find a derivation because there indeed is no derivation, or because of lack of skill on our part. The Truth Table method and the Truth Tree method give us ways of determining that an inference is valid, if it is valid, or invalid if it is invalid.

3. In addition, the truth table and truth tree methods are purely mechanical methods, which do not rely on any ingenuity on our part. However, the truth table method is artificial in that it does not demonstrate the steps in reasoning by which the conclusion is reached. The truth tree method solves this problem to some extent.

4. Another disadvantage of the truth table method, but less so of the truth tree method, is that truth tables can also be quite cumbersome to produce. A shorter version of the truth table method, called the targeted truth table method, is also given.

5. Both the truth table and truth tree methods are used to evaluate inferences whose propositions are in symbolic language.  

2 Truth Values & Truth Tables For The Logical Operators

1. A simple proposition can take one of two truth values, true (T) or false (F). (It's a little strange to say, but 'false' is a 'truth value'.) Whether a well-formed complex proposition is true or false depends solely upon whether the simple propositions in it are true or false, according to the following four tables:

~ S        S v T        S & T        S ⊃ T T T T        T T T        T T T

F T        T T F        T F F        T F FT F        F T T        F F T        F T T

F F F        F F F        F T F

The tables are read like this:  Tilde: If "S" (i.e. any well-formed proposition) is true — represented

by the non-bold letter "T" directly under the "S" — then "~S" is false — the

bold "F" directly under the tilde. And, in the second row, if "S" is false, then "~S" is true.  

Wedge: If "S" is true and "T" is true — the non-bold "T"s directly under "S" and "T" — then "S v T" is true — the bold "T" directly under the wedge. In the second row: If "S" is true and "T" is false then "S v T" is true. In the third row: if "S" is false and "T" is true then "S v T" is true. In the fourth: if "S" is false and "T" is false then "S v T" is false.  

Ampersand: First row: if "S" is true and "T" is true then "S & T" is true. Second row: if "S" is true and "T" is false then "S & T" is false. Third row: if "S" is false and "T" is true then "S & T" is false. Fourth: if "S" is false and "T" is false then "S & T" is false.  

Horseshoe: First row: if "S" is true and "T" is true then "S ⊃ T" is true. Second row: if "S" is true and "T" is false then "S ⊃ T" is false. Third row: if "S" is false and "T" is true then "S ⊃ T" is true. Fourth: if "S" is false and "T" is false then "S ⊃ T" is true.  

2. The truth table for tilde (negation) is obvious, given the assumption that there are two and only two truth values, true and false. If the proposition "Salt is an ingredient in ketchup." is true, then the proposition "Salt is not an ingredient in ketchup." is false. And if the proposition "Salt is an ingredient in ketchup." is false, then the proposition "Salt is not an ingredient in ketchup." is true.

The truth table for wedge (disjunction) indicates that a proposition is false when both disjuncts are false (in the fourth row); otherwise (in the other three) it is true. This truth table allows that a disjunction is true when both disjuncts are true. For example, the sign at a charity book sale "Everyone is welcome to donate goods or buy them." does not prohibit anyone from doing both. This interpretation is called "inclusive-or". Propositions which disallow the 'both' option, such as "you can have ice-cream or chocolate but not both." or "Jack will take care of Jim or else Gill will.", are called "exclusive-or". These can be translated into symbolic as "(S v T) & ~(S & T)".  

The truth table for ampersand (conjunction) shows (first row) that a conjunction is true only when both conjuncts are true; otherwise it is false. For example, "Jack is in Baghdad and Gill is in Boston." is false if one or both of the conjuncts is false.

The truth table for horseshoe (conditional) shows that a conditional is false only when the antecedent is true and the conditional is false (second row); otherwise it is true (rows one, three, and four). Rows one and two are perhaps obvious. Consider the following example: "If I take the full course of antibiotics, the infection will clear up.". This complex proposition is true if I take the full course and the infection clears, and is false if I take the full course of antibiotics and the infection persists. Rows three and four require some explanation. These last two lines both have a false antecedent, and also state that the conditional is true, regardless of the truth or falsity of the consequent. If I do not take the full course of antibiotics, is the conditional

false? No, but one might think that it is not true, either. Why should we default to "T"?

"If …, then …" is often understood causally, as in the example of the antibiotic and the infection. But "If …, then …" has other uses in addition to expressing causal connection, such as to express logical connections ("If "a ⊃ b" and "a", then "b"."), or connections based on definitions ("If it's water, it has hydrogen."). Conditionals are also used to speculate about what might be, as in "If I were to win the Lotto, I would quit my job.". (One can even use "If …, then …" to strongly assert the falsity of an antecedent, as in "If United loses on Saturday, then I'll eat my hat.").

All of these uses have it in common that the conditional is false if the antecedent is true and the consequent false. It might thus be better to think of "S ⊃ T" not as "If S, then T" but as "not both S and not T", or, in symbolic, "~(S & ~T)". In other words, that the conditional is false when the antecedent is true and the consequent is false is equivalent to saying that it can't simultaneously be that the antecedent is true and the consequent is false. When we think of the horseshoe in this way, we get the same results as in the truth table for horseshoe, above. If "S" is true and "T" is false (i.e. the second row), "~T" is true and "S & ~T" is true, and ~(S & ~T) is false. Under all other assignments (T and T, F and T, F and F), the proposition is true. And note that when "S" is false (as it is in the bottom of two lines of the truth table for the horseshoe), the internal conjunction is false, and the negation is true. Thus, "S ⊃ T" is equivalent to "~(S & ~T)".

"S ⊃ T" and "~(S & ~T) are, further, equivalent to "~S v T", as in "Either I do not take the full course of antibiotics or the infection will clear up.". Again, when "S" is true and "T" is false, the proposition is false, and when "S" is false (as it is in the bottom two lines of the truth table for the horseshoe), "~S" is true, and this is enough to make the disjunction true.

Understood in this truth-functional way, the horseshoe is called material conditional or material implication, since it is agnostic about the type of connection being asserted in the original English proposition and merely expresses the implication between two propositions.

3. Since the four types of complex proposition are composed of simple propositions and the four logical operators, we can work out the truth value of a proposition based on the truth values of the simple propositions and the four basic truth tables, given above. Consider the following proposition:

((p ⊃ q) & r) v ~q  T   F    T     F

Suppose that the truth values of the simple propositions are as follows: "p" is true, "q" is false, and "r" is true. What is the truth value of the proposition as a whole? The trick to determining the truth value of a complex proposition is to work from the inside(s) out. By "the insides" we mean most heavily nested operator(s), the operator(s) buried within the greatest number of pairs of parentheses. In this proposition, the horseshoe

has two pairs of parentheses around it, the ampersand has one, the negation has one (this pair of parentheses is invisible because of the rule that a tilde negates whatever follows it immediately!), and the wedge has none. Thus we will begin with the horseshoe and then move to the ampersand. At this stage, we will have a value for the left-hand disjunct. In the right-hand disjunct, we must deal with the negation. Finally, we can work out the truth value for the whole proposition by turning to the wedge, which is the main operator.

Beginning with the horseshoe and "p ⊃ q", since we are supposing for the sake of example that "p" is true and "q" is false, "p ⊃ q" is false; the second row of the truth table for horseshoe tells us this: propositions having the form "S ⊃ T" are false when "S" is true and "T" is false. This value (F, in bold, below the horseshoe) now stands for the proposition "p ⊃ q".

((p ⊃ q) & r) v ~q  T F F    T     F

Moving to the ampersand, "p ⊃ q" is the left-hand conjunct and its value is F, as we have just worked out. Given that "r" is true, "(p ⊃ q) & r" is false, since propositions having the form "S & T" are false when "S" is false and "T" is true. This value (F, in bold, below the ampersand) now stands for the proposition "(p ⊃ q) & r".

((p ⊃ q) & r) v ~q  T F F  F T     F

Moving to the disjunction, we see that the right-hand disjunct is a negation and we must deal with the negation before the disjunction. Since the proposition being negated "q") is false, it is true, for propositions having the form "~S" are true when "S" is false.

((p ⊃ q) & r) v ~q  T F F  F T    TF

Finally, we put the values for the left-hand and right-hand disjuncts together in the disjunction. "((p ⊃ q) & r) v ~q" is true, for propositions having the form "S v T" are true when "S" is false and "T" is true.

((p ⊃ q) & r) v ~q  T F F  F T  T TF

This means that when "p" is true, "q" is false and "r" is true, the complex proposition "((p ⊃ q) & r) v ~q" is true.

Exercises

For each proposition, underline "T" if it is true and "F" if it is false, when "b" is true, "c" is false, "d" is true, "e" is false, and "f" is true.

b c d e fT F T T T

Sampled ⊃ (b & ~c)T    T    FT    T   TFT    T T TFT T  T T TF

T   F

(1)    ~~b 

T   F

(2)    c ⊃ d

T   F

(3)    (e v ~f) ⊃ d

T   F

(4)    ~((e & ~f) ⊃ b)

T   F

(5)    f ⊃ (b ⊃ e)

T   F

(6)    ~(b ⊃ ~c) & (e ⊃ d)

T   F

(7)    ~f v (((b ⊃ c) ⊃ (~f & ~e)) v ~~d)  T   F

(8)    (~b v (b ⊃ c)) ⊃ ((~f & ~e) v ~d)  T   F

Answers to Even Numbers

For each proposition, underline "T" if it is true and "F" if it is false, when "b" is true, "c" is false, "d" is true, "e" is false, and "f" is true.

(2)    c ⊃ d

c ⊃ dF T T

T (4)    ~((e & ~f) ⊃ b)

~((e & ~f) ⊃ b)      F    T    T  

   F   FT    T   F F FT    T   F F FT  T TF  F F FT  T T

   F

(6)    ~(b ⊃ ~c) & (e ⊃ d)

~(b ⊃ ~c) & (e ⊃ d)     T    F     F   T       T   TF     F   T     T T TF     F   T   F T T TF     F   T   F T T TF     F T T   F T T TF  F  F T T

   F

(8)    (~b v (b ⊃ c)) ⊃ ((~f & ~e) v ~d)

  (~b v (b ⊃ c)) ⊃ ((~f & ~e) v ~d)    T    T   F        T    F     T   FT    T   F       FT   TF    FT   FT    T F F       FT   TF    FT

   FT F  T F F       FT   TF    FT   <----   FT F  T F F       FT F TF    FT   FT F  T F F       FT F TF  F FT   FT F  T F F   T   FT F TF  F FT

T   

(In fact, we can stop after the fourth line: once the antecedent is F, the conditional must be T)

3 Setting Up Truth Tables

1. In the previous section, we gave each of the simple propositions a truth value and used these to work out the value of a complex proposition. When we use The Truth Table Method, we do not know the values of the simple propositions and so imagine all of the possible combinations of values (also called assignments) and we work out the value of a complex proposition for each assignment. Moreover, we have to do this for all of the propositions in an inference. Luckily, there is a simple way to do both of these at once, called a truth table.

2. In this section, we present a procedure for setting up a truth table which will ensure that (a) no combination of truth values is duplicated and (b) no possible combination is left out.

First, determine how many rows the table will have. You can do this once you have put the passage into standard form using symbolic. Count up how many distinct simple propositions there are in the passage. Even if a letter appears more than once, count it only once. The following for example, involves only three distinct letters, even though "p" and "r" each occur twice.

1.    (p v q) ⊃ r2.    p    --3.    r

If the propositions of an inference involve only 1 distinct letter, then the truth table has 2 rows. If they involve 2 distinct letters in it, then the truth table has 4 rows. If 3 letters, 8 rows. If 4 letters, 16 rows. And so on. In general, if an inference involves n letters, then its truth table has 2n rows. Suppose that the propositions of an inference involve 3 letters:

p   q   r

Given that it has 3 letters, there will be 8 rows in the truth table, as 2 3

= 8. Second, fill in each column, as follows: We begin with the first simple

proposition and under it write down (n/2 = 8/2 =) 4 Ts and then 4 Fs.

The table at this stage looks like this:p   q   r

TTTTFFFF

For the next column, we divide by 2 again to get (4/2 =) 2 and so, below the second proposition letter we alternate between pairs of Ts and Fs.

p   q   r

T   TT   TT   FT   FF   TF   TF   FF   F

Dividing by 2 again to get (2/2 =) 1, under the third letter we alternate between single Ts and Fs. The complete assignment appears as follows:

p   q   r

T   T   TT   T   FT   F   TT   F   FF   T   TF   T   FF   F   TF   F   F

In general terms, the procedure runs as follows. First, determine the number of rows in the truth table. Let that number be n. Second, give a column of n/2 Ts under the first letter, and follow this with an equal number of Fs. At this point, the first column is done. Third, give a column of Ts under the second letter half as long as the column of Ts under the first, give a column of Fs under the Ts equal in length, give a column of Ts under the

Fs equal in length, and give a column of Fs under the Ts equal in length. At this point, the second column is done. And so on.

3. Once we have made sure that we will be considering every possible assignment of truth and falsity to the simple propositions involved, we can work out what the value of each whole proposition is under each assignment.

4 The Truth Table Method

1. The truth table method works like this: First, convert the inference into symbolic and put it in standard form. Second, make a truth table for the inference. Third, check the table for rows in which all of the premises are true and the conclusion is false. If there is such a row (or rows) the inference is invalid; if not, it is valid.

The truth table method rests on the fact that an inference is valid if it is impossible for the premises to be true and the conclusion false. If the truth table shows that it is possible for the premises to be true and the conclusion false, the inference is not valid. If this combination of true premises and false conclusion is impossible, the inference is valid.

2. Consider the following inference:Jones owns a Ford. So, either Jones owns a Ford, or Brown is in Barcelona.

With "o" standing for the proposition "Jones owns a Ford.", with "b" standing for "Brown is in Barcelona." in standard form, using symbolic, the inference is written as follows:

1.    o    --2.    o v b

In the truth table method, we make a truth table, as described above, for all of the propositions in the inference at once. We begin by picking out the simple propositions. This inference involves two simple propositions, "o" and "b". On one line, we first write down the simple propositions and then the premises and the conclusion. In this case, there is one premise and the conclusion. Thirdly, we generate all of the possible truth assignments to the simple propositions. Since there are two simple propositions, the table has four rows. The final step is to work out the truth values of the propositions of the inference, according to the basic rules and the procedure above. The final truth table looks like this:

o   b      o    o  v  b

T   T      T       TT   F      T       T

F   T      F       TF   F      F       F

(If you find it helpful, you can fill in the values for each proposition everywhere it appears. In this inference, you could fill them in under "o" and "b" in "o v b". The final truth table would look like this rather than the one just above:

o   b      o    o   v  b

T   T      T    T   T  TT   F      T    T   T  FF   T      F    F   T  TF   F      F    F   F  F                    )

We put the column of values under each of the propositions in bold (or when writing by hand, draw a box around it). For complex propositions, this column should be under the main operator. In this inference, it appears under the wedge in the conclusion.

Finally, we inspect the table in search of a row in which the premise is true and the conclusion is false. If we find such a row, the inference is invalid, since we know from our earlier discussion of the concept of 'validity' that no inference with true premises and a false conclusion can be valid. The premise is true in the first and second rows, and so is the conclusion. In the third and fourth rows, the premise is false. So there is no row in which the premise is true and the conclusion is false and, thus, this inference is valid.

2. Here is a truth table for an instance of MA, the inference "p ⊃ q. p. So, q.". The first step is to identify the simple propositions. In this case, there are two: "p" and "q". We write these down in a row. The next (second) step is to write down the propositions of the inference, on the same line as the simple propositions. After the second step, the truth table looks like this:

p   q        (p  ⊃  q)            p    qThe third to step is to generate the possible combinations of assignments of T and F to the simple propositions. Since there are two simple propositions, there are four possible assignments of truth values. After step three, the truth table looks like this:

p   q        (p  ⊃  q)        p     qT   TT   FF   TF   F

The fourth step is to work out the truth values of the propositions for each assignment, as described above. After this fourth step, the table looks like this:

p   q        (p  ⊃  q)        p    qT   T           T             T    TT   F           F            T    FF   T           T            F    TF   F           T           F    F

There is no row in which the premises are true and the conclusion false. Thus, this inference is valid.

3. In contrast to the two examples so far, this inference is invalid:1.    b    --2.    b & c

Its truth table looks like this:b   c      b    b  &  c

T   T      T       TT   F      T       F    <----F   T      F       FF   F      F       F

In the second row, the premise is true while the conclusion is false. We point to any invalidating row with an arrow.

4. Now consider the following inference:If Henry was not required to work and was able to save up enough money, he will be at the concert tonight.  He's not at the concert. So, he must not have been able to save up the money.

With "w" standing for "Henry was required to work.", "s" standing for "Henry was able to save up enough money.", with "c" standing for "Henry is at the concert." in standard form, using symbolic, it looks like this:

1.    (~w & s) ⊃ c2.    ~c    ---------------3.    ~s

This inference involves three distinct propositions, and so its truth table has eight rows. The completed table looks thus:

w   s   c      (~w  &  s)  ⊃   c    ~c    ~sT   T   T       F   F      T        F     FT   T   F       F   F      T        T     F     <----T   F   T       F   F      T        F     TT   F   F       F   F      T        T     TF   T   T       T   T      T        F     FF   T   F       T   T      F        T     FF   F   T       T   F      T        F     TF   F   F       T   F      T        T     T

In the second row, the premises are true while the conclusion is false. The inference, thus, is invalid.

Exercises

SampleUse the truth table method to show that some arbitrary instance of CC is valid.

1.    a ⊃ b         a b  a ⊃ b ~b ~a2.    ~b          T T    T   F  F

------         T F    F   T  F3.    ~a          F T    T   F  T

F F    T   T  T

No line has true premises and a false conclusion. So, the inference is valid.

(1) Use the truth table method to show that some arbitrary instance of Chain is valid.

(2) Use the truth table method to show that some arbitrary instance of MC is not valid.

(3) Use the truth table method to show that some arbitrary instance of CA is not valid.

(4) Use the truth table method to show that some arbitrary instance of Conj. is valid.

(5) Use the truth table method to show that some arbitrary instance of CD is valid.

(6) Use the truth table method to determine whether the following inference is valid or not valid.

1.    (b v ~c) ⊃ d2.    ~c

-------------

3.    d

(7) Consider the following inference:

If national elections deteriorate into television popularity contests, smooth-talking morons will get elected. So clearly, smooth-talking morons won't get elected if the elections don't deteriorate into television popularity contests.

(a) Make a translation key for it.(b) Relative to the key, translate the inference into symbolic and put it in

standard form. (c)  Relative to what you have in standard form, make a truth table for the

inference. (d) Relative to your truth table, is the inference valid or not valid?

Answers to Even Numbers

(2)    Use the truth table method to show that some arbitrary instance of MC is not valid.

1.    a ⊃ b2.    b

------3.    a

a    b        a ⊃ b    b        aT    T         T      T        TT    F         F      F        TF    T         T      T        F <----F    F         T      F        F

(4)    Use the truth table method to show that some arbitrary instance of Conj. is valid.

1.    a2.   b

--3.   a & b

a    b         a      b      a & bT    T         T      T        TT    F         T      F        FF    T         F      T        FF    F         F      F        F

(6)    Use the truth table method to determine whether the following inference is valid or not valid.

1.    (b v ~c) ⊃ d2.    ~c

-------------3.    d

b    c    d   (b v ~c)⊃ d     ~c        dT    T    T     T F  T       F         TT    T    F     T F  F        F        FT    F    T     T T  T        T        TT    F    F     T T  F        T        FF    T    T     F F  T        F        TF    T    F     F F  T        F        FF    F    T     T T  T        T        TF    F    F     T T  F        T        F

Valid – there is no line with T premises (the horseshoe in the first proposition and the tilde in the second) and F conclusion

5 Logical Equivalence & Inequivalence, & Logical Contradiction1. The proposition "If Jones is in Columbus, then he is in Ohio." is

logically equivalent to "If Jones is not in Ohio, then he is not in Columbus.". Truth tables bear this out—the two tables have the same column of final values. With "c" standing for "Jones is in Columbus." and with "o" standing for "Jones is in Ohio.", the two become:

c  ⊃  o                ~o  ⊃  ~c

A truth table for both propositions simultaneously looks like this:c   o      c  ⊃  o        ~o  ⊃   ~cT   T         T           F   T   FT   F         F           T   F   FF   T         T           F   T   TF   F         T           T   T   T

There is no row in which the two propositions have different truth values. So, they are logically equivalent.

2. In contrast, b ⊃ c and c ⊃ b are logically inequivalent. Consider the following table:

b   c      b  ⊃  c    c  ⊃  b

T   T         T          TT   F         F          TF   T         T          FF   F         T          T

Not every row has the same ultimate value; rows 2 and 3 both have opposite truth values. 3. If two propositions have opposite values on every line, they are said to be logical contradictories.

Exercises

(1)    Make a truth table for the propositions "p" and "~~p" and, relative to your truth table, determine whether they are logically equivalent or logically inequivalent.

(2)    Make a truth table for the propositions "~(b & c)" and "~b v ~c", and, relative to your truth table, determine whether they are logically equivalent or logically inequivalent.

(3)    Make a truth table for the propositions "~(b & c)" and "~b & ~c" and, relative to your truth table, determine whether they are logically equivalent or logically inequivalent.

(4)    Make a truth table for the propositions "b ⊃ c" and "~b v c", and, relative to your truth table, determine whether they are logically equivalent or logically inequivalent.

(5)    Make a truth table for the propositions "p ⊃ (q ⊃ r)" and "(p & q) ⊃ r" and, relative to your truth table, determine whether they are logically equivalent or logically inequivalent.

(6)    Make truth tables for the sentences "a & (b v c)" and "(a & b) v (a & c)" and, relative to your truth table, determine whether they are logically equivalent or logically inequivalent.

6 Targeted Truth Tables

1. Truth tables can be cumbersome, especially when there are multiple simple propositions. One way to shorten the process is by using the targeted truth table method. This method relies on the fact that in the truth table method we are interested in lines of the truth table in which the premises are true and the conclusion is false. The targeted truth table method attempts to find an assignment (or assignments) on which the premises are true and the conclusion is false, without making a full truth

table. If such an assignment(s) can be found, the inference is invalid; if not, the inference is valid.

2. Consider the following inference, in standard form using symbolic: 1.    b ⊃ c2.    a & (c v d)     ------------3.    d v b

Since there are four distinct simple propositions, the full truth table would have 16 lines. We can shorten the process by using the targeted truth table method. We begin by writing the simple propositions, the premises and conclusion on a line, as follows:

a b c d    b ⊃ c      a  &  (c v d)       d v b

Since we are targeting the row(s) of the table on which the conclusion is false and the premises true, we think about which assignment(s) would make the conclusion false, or any of the premises true. In this example, the conclusion makes a good place to begin, since a disjunction is false only when both disjuncts are false. So we assign F to both "d" and "b" in the conclusion, and everywhere else either "d" or "b" appears in the premises. We fill in these values for "b" and "d" wherever they appear and we write an "F" under the wedge in the conclusion. At this stage, the targeted truth table looks like this:

a b c d    b ⊃ c      a  &  (c v d)       d v b  F   F    F                     F        F F F

Using this assignment for "d" and "b", can an assignment be found for the remaining letters ("a" and "c") such that the premises are true? The first premise is a conditional, with a false antecedent, so it will be true regardless of the value we give to "c". We can thus leave the first premise, and the value of "c", aside for the time being. The second premise is a conjunction. In order for it to be true both conjuncts must be true. So we assign T to "a" and T to "c".

a b c d    b ⊃ c      a  &  (c v d)       d v bT F T F    F T T      T  T   T T F        F F F

We have thus shown that the inference is invalid, because when "a" is T, "b" is F, "c" is T and "d" is F, the premises are true and the conclusion false.

3. Consider the following, different, inference:

1.    b & c

2.    a ⊃ (c v d)     ------------3.    d v b

Employing the same procedure as before, we write out the propositions involved, and assign F to "d" and "b" in order to make the conclusion F. This time however, we can see not only that the conclusion is F (we write an "F" under the wedge in the conclusion) but also that the first premise will be F (even without knowing the value of "c". So, we can write an "F" under the ampersand.

a b c d    b & c     a ⊃ (c v d)       d v b  F   F    F F                F        F F F

Since the first premise cannot be true when we make the conclusion false (F) by assigning F to both b and d, (because for a conjunction to be true, both conjuncts must be true) this inference is valid; it is impossible for the premises to be true and the conclusion false.

4. In general, we start by looking for assignments that we must make in order to make either a premise true or the conclusion false; these will give us fixed pieces of information. We can immediately assign values to any propositions that are not complex. If a premise is a single proposition letter, assign T to it; if a conclusion is a single proposition letter, assign F to it. After this, it is a good idea to begin with the conclusion, since there is only ever one conclusion and it must be made false. Moreover, propositions which are either negations, disjunctions or conditionals are false under only one assignment. A negation is false when what is negated is true; a disjunction is false when both disjuncts are false; and, a conditional is false only when the antecedent is true and the consequent is false.

5. However, if the conclusion is a complex statement using parentheses, the best strategy might not be to begin with the conclusion. For example, if the conclusion is "~(a ⊃ b)", we know that this is false when "a ⊃ b" is true. But there are multiple assignments under which "a ⊃ b" is true. Conjunctions, similarly, are false when either, or both, of the conjuncts are false.

In these cases, we should look to the premises. For example, consider the following inference:

1.    b & c2.    b v d3.    d ⊃ a    -------4.    ~(a ⊃ b)

When we write out the propositions on a line, we see that there is no single assignment under which the conclusion is false. When we look at the premises, we are confronted with a conjunction, a disjunction and a conditional. Conjunctions, which are false under multiple assignments, are true only when both conjuncts are true, so the premise to begin with is the first one: we assign T to both "b" and "c" and fill in what else we can:

a b c d    b & c      b v d     d ⊃ a       ~(a ⊃ b)  T T      T T T      T T                   F   T T

The second premise is already true, so we leave the value of "d" in the second premise aside for the moment. We are not forced into making an assignment to "d" based on the third premise either, since if "a" were true, it wouldn't matter what the value of "d" was. What about "a"? Again, we are not forced into a value for "a". Based on the third premise, it could be either, since "d" could be either; based on the conclusion, it could be either, since "b" is T. So there are a number of assignments which would show this inference to be invalid. If we assign T to "a", "d" can be either T or F, as follows:

a b c d   b & c      b v d     d ⊃ a       ~(a ⊃ b)T T T     T T T      T T         T T       F T T T

Alternatively, if we assign F to "d", "a" can be either T or F, as follows: a b c d    b & c      b v d     d ⊃ a       ~(a ⊃ b)  T T F    T T T      T T F     F T         F   T T

Any one of the following assignments of values is sufficient to show that the inference is invalid:

a b c d T T T TT T T F F T T F

6. Sometimes there is no proposition which forces us into assigning truth values to any of the proposition letters involved. In such cases we must try out different assignments one at a time. Consider the following such inference:

a b d    b v d     d ⊃ a     b & aIn this inference we have a conclusion which is a conjunction together

with premises which are a disjunction and a conditional. We are not forced to assign T or F to any letter. Instead, we must work through the various possible assignments. To make the conclusion false, three assignments are possible: both "b" and "a" are false, or, "b" alone is false, or, "a" alone is

false. We can work through each  of these. If none of them allow for true premises and a false conclusion, the inference is valid.

a b d    b v d     d ⊃ a     b & aF F      F             F     F F FT F      F             T     F F TF T      T             F     T F F

Making both false (on the first line), we see that "d" must be F if the second premise is to be true, but when we do that, we see that the first premise is false. We get:

a b d    b v d     d ⊃ a     b & aF F F    F F F     F T F     F F FT F      F             T     F F TF T      T             F     T F F

Note that the failure of this assignment does not show that the inference is valid. Validity means that no assignment results in true premises and a false conclusion; all we have shown so far is that we do not get true premises and a false conclusion when we assign F to "a" and "b". Since this was only one of the assignments that would make the conclusion false, we must try the others.

We know that in order to make the conclusion false, either "b" or "a" must be false, so let us consider the second possibility, which assigns F to 'b' and T to "a":

a b d    b v d     d ⊃ a     b & aF F F    F F F     F T F     F F F T F      F           T T     F F TF T      T             F     T F F

In order to make the first premise true, "d" must be T, and now we have found an assignment which shows that the inference is invalid, one which makes the premises true and the conclusion false:

a b d    b v d     d ⊃ a     b & aF F F    F F F     F T F     F F F T F T    F T T     T T T     F F TF T      T             F     T F F

At this point we can stop, since any assignment which yields true premises and a false conclusion shows that the inference is not valid. There might be additional assignments which show that the inference is invalid, but one is sufficient to show that it is not valid. (The third assignment also demonstrates invalidity, in fact.)

To repeat, the crucial point of this example is that the fact that the first assignment we tried did not show the inference to be not valid was not

sufficient to show that it was valid. To show that an inference is valid we would need to try all of the possible assignments that will either make the conclusion false or make one of the premises true.

Exercises

Part 1. For each inference, use a Targeted Truth Table to determine whether the inference is valid or not valid.

Sample

1.    e ⊃ f2.    ~(e & f)    ---------3.    f

            e    f          e ⊃ f       ~(e & f)        f       F            F            F         F (step 1)     F    F           F T F      T F F F         F (step 2)

Not valid, when e is F and f is F.

(1)    

1.    p ⊃ a2.    ~(a v s)     ---------3.    ~p

(2)

1.    t ⊃ s2.    s ⊃ b     ------3.    t ⊃ b

Part 2. For each inference, make a translation key, put the inference in standard form using symbolic and use a Targeted Truth Table to determine whether the inference is valid or invalid.

(3)         If we paint the plant with soapy water, the aphids will disappear. But it's not the case that either the aphids will disappear, or the spider mites. So we will paint the plant with soapy water.

(4)    If Jack gets more training, he qualifies for the Special Ops unit. If Jack qualifies for the Special Ops unit, he is shipping to Baghdad. So, if Jack is shipping to Baghdad, he gets more training.

(5)    If Jack gets more training, he will qualify for the Special Ops unit. If Jack qualifies for the Special Ops unit, he will immediately be shipped

to Baghdad. So, either Jack gets more training or he will not immediately be shipped to Baghdad.

Answers to Even Numbers

(2)   

1.    t ⊃ s2.    s ⊃ b     -------3.    t ⊃ b

    t    s    b            t ⊃ s             s ⊃ b       t ⊃ b  T        F            T                    F       T F F  T        F            T T T             T F F       T F F

Valid - It is impossible to assign values which make the inference invalid

(4)         If Jack gets more training, he qualifies for the Special Ops unit. If Jack qualifies for the Special Ops unit, he is shipping to Baghdad. So, if Jack is shipping to Baghdad, he gets more training.

1.    t ⊃ q2.    q ⊃ b    ------3.    b ⊃ t

t    q    b            t ⊃ q           q ⊃ b       b ⊃ tF         T            F                    T       T F FF    T    T            F T T            T T T       T F F

Invalid - when "t" is F, "q" is T and "b" is T, the premises can be made true and the conclusion false

7 The Truth Tree Method

1. Truth trees, like truth tables, will show that an inference is valid if it is valid, or not valid if it isn't. They have the advantage of dealing more briefly than truth tables with inferences involving a number of simple propositions. Their spatial/graphical nature also makes them more intuitive for some people.

2. Truth trees are like targeted truth tables. We assume the premises are true and the conclusion is false. If the inference is valid, we have just assumed a contradiction, and this contradiction will show itself when we

look at the simple propositions. On the other hand, if the simple propositions can be consistent, the inference is invalid.

3. We begin the process by writing the premises and the negation of

the conclusion in a single vertical column. In the truth tree method we do not use "T" and "F" as contradictory truth values; rather, we use the simple proposition letters and their negations. Thus, to negate the conclusion is to assume that the conclusion itself is false.

This initial vertical column can be thought of as the trunk of the tree starting to grow. The tree is growing downwards, by the way. It is upside-down. (Perhaps this method should be called the "roots" method!)

For example, a simple Elim. inference would be set up like this:a v b~a~b

Note that the conclusion "b" has been negated.

4. The next step is to "decompose" any complex propositions listed, except negations of simple propositions. To decompose a proposition, we strike it out and extend each branch of the tree downwards in accordance with the rules of decomposition. A full list of the rules of decomposition will be given shortly.

In our Elim. example, neither the second nor the third line need to be decomposed, since they are both negated simple propositions. Only the first line needs to be decomposed. The rule for decomposing a disjunction is to "branch" into the left-hand disjunct and the right-hand disjunct. (A complete set of rules is given below.) We strike out the line decomposed.

a v b~a~b

   a     b

5. After each decomposition, check for contradictions along each branch, including the trunk. In our example, moving up from the bottom of the left-hand branch, we see that it contains a contradiction between "a" on line 4 and "~a" on line 2, and the right-hand branch contains a contradiction between "b" on line 4 and "~b" on line 3. We place an "X" below closed branches.

a v b~a~b

a    bX    X

6. Our Elim. example has closed all of its branches. It is therefore valid. If there were any unclosed branches, we would continue the process of decomposition, one proposition at a time, until all complex propositions have been decomposed. If all of the propositions have been decomposed and there are still unclosed branches, the inference is not valid.

7. Some decomposition rules are branching and some are non-

branching. By "branching" we mean that the proposition decomposes horizontally, indicating that there are two ways in which the proposition could be true, while "non-branching" means that the proposition decomposes vertically, indicating that the truth of the proposition requires that both parts are true.

Conjunctions, for example, decompose vertically. Consider the inference from "a & b" to "a". We would set this up as:

a & b~a

Only the first line needs to be decomposed. Conjunctions decompose vertically, extending the trunk of the tree. After decomposing line 1 and looking for contradictions, our example looks as follows:

a & b~aabX

There is only one "branch" (in fact, the tree is just a trunk) and it contains a contradiction, between "a" on line 3 and "~a" on line 2. The inference is valid.

8. At the end of the process, either every branch will be closed, or not every branch will be closed. If every branch is closed, we have shown that the assumption of invalidity was contradicted, and so the inference is valid. Each branch that remains open, however, illustrates a counter-example to the validity of the inference, which shows the inference to be invalid.

Here is a demonstration of the invalidity of MC. Letting "S" and "T" be "a" and "b", the propositions involved are "a ⊃ b", "b", "a", and so we write down:

a ⊃ bb~a

There is only one proposition requiring decomposition ("a ⊃ b"), which decomposes by branching into "~a" and "b". We add these branches to the

trunk on the next line below and strike out the proposition being decomposed:

a ⊃ bb~a

~a     b

We check each branch for contradictions. There are none. It is possible for the premises to be true and the conclusion false, when "~a" is true (or, "a" is false) and "b" is true. The inference is declared not valid.

9. Here is the complete set of rules of decomposition. With "S" and "T" standing for any proposition, whether simple or complex, the non-branching rules are as follows:    S & T        ~(S v T)        ~(S ⊃ T)        ~~S       S             ~S           S         S        T             ~T          ~T   

and the branching rules are as follows:~(S & T)             S v T               S ⊃ T

  ~S         ~T        S         T        ~S         T

Remember that as you decompose each proposition, you strike it out in your tree.

Some of these rules will be familiar from the truth tables above. "S & T" decomposes into a non-branching "S" and "T" because "S & T" is true when both "S" and "T" are true. (Compare with the truth table for "S & T", above.) "S v T" branches into "S" and "T" because a disjunction is made true when either one of the disjuncts is true. (Again, compare the truth table for "S v T" above.) "S ⊃ T" branches because a conditional is true when either the antecedent is not true or the consequent is true. (Again, compare the truth table for "S ⊃ T" above.) The rules for "~~S" can be understood by comparison with the truth table in section 2 and the rule of double negation from the Method of Derivation (see 6.1 of that chapter) and "~(S & T)" and "~(S v T)" can be understood by comparison with De Morgan's rule (in 6.6.)

10. Decompose non-branching propositions before branching ones. This is because, if a tree already has branches, any further decompositions must be applied to all of the branches below it. We can prevent some unnecessary writing by extending the trunk first, since the trunk is included in all branches. Consider the following inference, in standard form:

1.    a v ~b

2.    ~a & c--------

3.    ~b

We set up the truth tree as follows, writing "~~b" on the third line as the negation of the conclusion.

a v ~b ~a & c ~~b

It's a good idea to decompose "~a & c" and "~~b" before the disjunction in the first line, because they are non-branching while it is branching. After decomposing the second line, we get:

a v ~b~a & c~~b~ac

We check for contradictions, but seeing none, proceed to decompose the third line ("~~b"). We cross out "~~b" on the third line and add "b" to the bottom of the tree, to get:

a v ~b~a & c~~b~acb

There are still no contradictions, so we proceed to decompose the first line:a v ~b~a & c~~b~acb

a    ~b

Again, we check for contradictions on each branch. The left-hand side contains a contradiction between the "a" on the final line and the "~a" on the fourth line. So, we place an "X" beneath the left-hand branch. The right-hand side contains a contradiction, between the "~b" on the final line and the "b" on the sixth line. So, we place an "X" beneath the right-hand branch.

a v ~b~a & c

~~b~acb

a    ~bX      X

Since we have finished decomposing, and there are only these two branches, both of which have been closed off, the inference is valid.

If we had decomposed line 1 first, the tree would have looked like this:

a v ~b~a & c~~b

a        ~b

Further decompositions (of line 2 and line 3) would then have to be written under both branches. After decomposing line 2:

a v ~b~a & c~~b

a        ~b~a        ~ac        c

X       

And after decomposing line 3:a v ~b~a & c~~b

a        ~b~a        ~ac        cX        b        X

11. Any unclosed branch is sufficient to show that the inference is not valid. In the following example, which is already in truth tree format, the first two lines are the premises, while the third is the negation of the conclusion.

a ⊃ f(a & c) v ~b

~~b

We begin by decomposing third line ("~~b") without branching, on line four. Both the first and second lines will cause the tree to branch. Taking the first line first, we get:

a ⊃ f(a & c) v ~b

~~bb

~a      f

We look for contradictions along each branch, and find none. The second line is now decomposed. The results of its decomposition must be placed at the bottom of each branch below it. We look for contradictions and find that the second and fourth branches show a contradiction, between "~b" and "b" on the fourth line.

a ⊃ f(a & c) v ~b

~~bb

~a             f a & c    ~b    a & c    ~b

          X              X   

Finally, we decompose "a & c":~a ⊃ f

(a & c) v ~b~~bb

~a             f a & c    ~b    a & c    ~b

                    a       X      a       X                    c              c

The first branch closes, since the "a" on the seventh line contradicts the "~a" of the fifth line:

~a ⊃ f(a & c) v ~b

~~bb

~a             f a & c    ~b    a & c    ~b

                    a       X     a        X                    c             c                    X

However, the third branch does not close, and so the inference is invalid, when "a" is true", "b" is true, "c" is true, and "f" is true.

Exercises

SampleUse the truth tree method to show that the following (an instance of CC) is valid:

1. a ⊃ b2. ~b

------3. ~a

a ⊃ b~b~~aa

  ~a       b  X       X

(1)    Use the truth tree method to show that the following (an instance of MC) is invalid:    

1. a ⊃ b2. b

-------3. a

(2)    Use the truth tree method to show that the following (an instance of Simp.) is valid:

1. a & b------

2. a

(3)    Use the truth tree method to show that the following (an instance of Elim.) is valid:

1. a v (b ⊃ c)2. ~a

------------3. b ⊃ c

(4)    Use the truth tree method to show that the following (an instance of DD) is valid:

1. a ⊃ b2. c ⊃ d3. ~b v ~d

----------4. ~a v ~c

(5)    Use the truth tree method to determine whether the following inference is valid or invalid.

1.    (b v ~c) ⊃ d2.    ~c

--------------3. d

(6)       Consider the following inference:

If national elections deteriorate into television popularity contests, smooth-talking morons will get elected. So clearly, smooth-talking morons won't get elected if the elections don't deteriorate into television popularity contests.

(a) Make a translation key for it.(b) Relative to the key, translate the inference into symbolic and put it in

standard form.(c)  Relative to what you have in standard form, make a truth tree for the

inference.(d) Relative to your truth tree, is the inference valid or invalid?

Answers to Even Numbers

(2)    Use the truth tree method to show that the following (an instance of Simp.) is valid:

a & b ~a   a b   X

(4)    Use the truth tree method to show that the following (an instance of DD) is valid:

a ⊃ bc ⊃ d

  ~b v ~d  ~(~a v ~c)

~~a

~~c a c

~b          ~d ~a    b    ~a       b  X    X     X    ~c   d                   X   X   

(6)         Consider the following inference:

If national elections deteriorate into television popularity contests, smooth-talking morons will get elected. So clearly, smooth-talking morons won't get elected if the elections don't deteriorate into television popularity contests.

(a) Make a translation key for it.d = National elections deteriorate into television popularity contests.e = Smooth-talking morons get elected.

(b) Relative to the key, translate the inference into symbolic and put it in standard form.

1.    d ⊃ e        ------2.    ~d ⊃ ~e

(c)  Relative to what you have in standard form, make a truth tree for the inference.

d ⊃ e ~(~d ⊃ ~e)      ~d

~~e  e

~d   e      

(d) Relative to your truth tree, is the inference valid or invalid?

Not valid – neither branch closes off.

8 A Summary Of Truth Conditions — Truth Tables & Truth Trees

Truth Tables

~ S        S v T        S & T        S ⊃ T T T T        T T T       T T T

F T        T T F        T F F        T F FT F        F T T        F F T        F T T

F F F        F F F       F T F

Truth Trees

Non-branching:

    S & T        ~(S v T)        ~(S ⊃ T)        ~~S          S             ~S            S          S         T             ~T           ~T   

Branching:

~(S & T)                S v T                 S ⊃ T ~S          ~T         S       T            ~S       T