﻿ 1.5 Truth-value analysis

### 1.5 Truth-value analysis

Truth tables show how exactly truth values of formulas depend on the truth values of their propositional letters. In particular, they serve to determine whether a formula is a tautology, a contradiction, or contingent. Truth-value analysis is an alternative method that serves the same purpose, but has some advantages.1

To see what this method is, let us ask ourselves what follows for the truth value of the formula “[p∧(¬qr)]∨(¬pq)” if one of its propositional letters – say “p” – gets a certain truth value. What happens, for example, if “p” has the value T? To answer this question, we replace in the formula “p” by “T”:

 [p∧(¬q∨r)] ∨ (¬p∧q) p: Т [Т∧(¬q∨r)] ∨ (¬Т∧q)

The conjunction of a true sentence and an arbitrary sentence always has the truth value of the latter sentence, i.e. T∧α has the true value of α. This is because if α has the value T, T∧α also has the value T (the conjunction of two true sentences is a true sentence), and if α has the value F, T∧α also has the value F (the conjunction of a true sentence and a false sentence is a false sentence). This principally gives us the right to replace expressions of the form T∧α by α. So in the expression above, we may replace “T∧(¬qr)” by “¬qr”:

 [p∧(¬q∨r)] ∨ (¬p∧q) p: Т [Т∧(¬q∨r)] ∨ (¬Т∧q) (¬q∨r) ∨ (¬Т∧q)

We have thus far obtained that when “p” has the value T, the initial formula has the same truth value as “(¬qr)∨(¬T∧q)”. In the last formula, we can replace “¬T” by “F” because the negation of a true sentence is a false sentence.

 [p∧(¬q∨r)] ∨ (¬p∧q) p: Т [Т∧(¬q∨r)] ∨ (¬Т∧q) (¬q∨r) ∨ (¬Т∧q) (¬q∨r) ∨ (F∧q)

“F∧q” may in turn be replaced by “F” since when one of the constituents of a conjunction is false, the conjunction is also false, regardless of the truth value of the other constituent (a conjunction is true only when both conjuncts are true). Generally, an expression of the form F∧α can be replaced by “F”. As a result of the substitution, we get that when “p” has the value T, the initial formula has the same true value as “(¬qr)∨F”:

 [p∧(¬q∨r)] ∨ (¬p∧q) p: Т [Т∧(¬q∨r)] ∨ (¬Т∧q) (¬q∨r) ∨ (¬Т∧q) (¬q∨r) ∨ (F∧q) (¬q∨r) ∨ F

The last expression is a disjunction with a false constituent. The disjunction has the truth value of the other constituent, i.e. H∨α (the order of the disjuncts is irrelevant) has the same true value as α. The reason is that when α is true, F∨α is also true (for a disjunction to be true, it suffices at least one of the disjuncts to be true), and when α is false, H∨α is also false (a disjunction of two false sentences is false). So, generally, an expression of the form H∨α may be replaced with α. In the particular case, we substitute “¬qr” for “(¬qr)∨Н”:

 [p∧(¬q∨r)] ∨ (¬p∧q) p: Т [Т∧(¬q∨r)] ∨ (¬Т∧q) (¬q∨r) ∨ (¬Т∧q) (¬q∨r) ∨ (F∧q) (¬q∨r) ∨ F ¬q∨r

We get that when “p” has the value T, the truth value of the whole expression is the same as that of “¬qr”. The truth value in question depends on the truth values of “q” and “r”. Therefore, we choose one of them, say “r”, and consider successively the cases when it is true and when it is false. First, for the case when “r” is true, we replace “r” in “¬qr” with “T”:

 [p∧(¬q∨r)] ∨ (¬p∧q) p: Т [Т∧(¬q∨r)] ∨ (¬Т∧q) (¬q∨r) ∨ (¬Т∧q) (¬q∨r) ∨ (F∧q) (¬q∨r) ∨ F ¬q∨r r: Т ¬q∨Т

When one constituent of a disjunction is true, the disjunction is also true, no matter what the truth value of the other disjunct is, i.e. an expression of the form T∨α always has the value T. This gives us the right to replace “¬q∨T” with “T”:

 [p∧(¬q∨r)] ∨ (¬p∧q) p: Т [Т∧(¬q∨r)] ∨ (¬Т∧q) (¬q∨r) ∨ (¬Т∧q) (¬q∨r) ∨ (F∧q) (¬q∨r) ∨ F ¬q∨r r: Т ¬q∨Т Т

We get that when “p” and “r” have the value T, the initial formula is true, regardless of the truth value of “q”.

For the other possible case, when “r” has the value F, we replace “r” in “¬qr” with “F”. As already seen, “F∨α” has the same true value as α, and since the order of the disjuncts is irrelevant “¬q∨F” may replaced by to “¬q”:

 [p∧(¬q∨r)] ∨ (¬p∧q) p: T [Т∧(¬q∨r)] ∨ (¬Т∧q) (¬q∨r) ∨ (¬Т∧q) (¬q∨r) ∨ (F∧q) (¬q∨r) ∨ F ¬q∨r r: Т r: F ¬q∨Т ¬q∨F Т ¬q

So when “p” is true and “r” is false, the initial formula has the truth value of “¬q”, i.e. it is false when “q” is true, and true when “q” is false:

 [p∧(¬q∨r)] ∨ (¬p∧q) p: Т [Т∧(¬q∨r)] ∨ (¬Т∧q) (¬q∨r) ∨ (¬Т∧q) (¬q∨r) ∨ (F∧q) (¬q∨r) ∨ F ¬q∨r r: Т r: F ¬q∨Т ¬q∨F Т ¬q q: Т q: F F Т

So far, the truth-value analysis gives us half of the information that is contained in the formula’s truth table, i.e. it shows its possible truth values when “p” is true (the other half is when “p” is false). Moreover, it is already clear that the formula is contingent. If we were only interested in the type the formula (whether it is a tautology, contradiction or contingent), we could stop here. This shows one of the advantages of truth-value analysis over truth tables. Since a table is made column by column rather than row by row (otherwise it is more difficult), to determine the type of a formula, we need to complete the whole table (only the last column shows what the type is), whereas using truth-value analysis we may get an answer without the analysis being complete. When we are interested in whether a formula is a tautology, if in the analysis we get to “F”, we can stop, as this gives a negative answer to this question.

The other half of the analysis concerning the case where “p” has the value F is obtained analogously:

 [p∧(¬q∨r)] ∨ (¬p∧q) p: Т p: F [Т∧(¬q∨r)] ∨ (¬Т∧q) [F∧(¬q∨r)] ∨ (Т∧q) (¬q∨r) ∨ (¬Т∧q) F ∨ (Т∧q) (¬q∨r) ∨ (F∧q) Т∧q (¬q∨r) ∨ F q ¬q∨r q: Т q: F r: Т r: F Т F ¬q∨Т ¬q∨F Т ¬q q: Т q: F F Т

The expression under “p: F” is obtained from the initial formula by replacing “p” with “F”. In the replacement of the second occurrence of “p”, we wrote directly “T” instead of writing “¬F” and then “T”. The next line is derived by replacing “F∧(¬qr)” with “F”. The reason is that (as we saw above) an expression of the form F∧α is equivalent to F. The next expression is derived by the (already introduced) rule according to which F∨α can be replaced by α – α in this case is “T∧q”. “T∧q” is then replaced by “q” by the (introduced) rule according to which T∧α can be replaced by α. As a result, we get that when “p” is false, the truth value of the initial formula does not depend on the truth value of “r” but is the same as that of “q”. This demonstrates the other advantage of true-value analysis over truth tables – often it is shorter because it provides similar “compressed” information.

We used the following replacement rules above:

 negation: ¬T is replaced by F ¬F is replaced by T conjunction: T∧α (or α∧T) is replaced by α F∧α (or α∧F) is replaced by F disjunction: T∨α (or α∨T) is replaced by T F∨α (or α∨F) is replaced by α

The rules regarding the conditional are as follows (in the case of conditional, unlike conjunction, disjunction and biconditional, the order of the constituent sentences is important, so here we have four, not two, rules):

 conditional: F→α is replaced by T α→T is replaced by T T→α is replaced by α α→F is replaced by ¬α

Where these rules come from can be seen through the conditional truth table:

 α β α → β Т Т Т Т F F F Т Т F F Т

The table shows that when the antecedent of a conditional is false (the last two rows), the conditional is true regardless of the truth value of its consequent. This gives us the right to replace F→α with T. The table also shows that when the consequent of a conditional is true (the first and the third raw), the conditional is true regardless of the truth value of its antecedent. This gives us the right to replace α→T with T. The reason for the third rule is the following. When α is true, T→α is also true because then the conditional connects two true sentences (the first row). When α is false, T→α is also false, because then the conditional has a true antecedent and a false consequent (second row). So, T→α always has the same truth value as α and can be replaced by it. With regard to the fourth rule, when α is true, α→F is false (the second row), and when α is false, α→F is true (the last row); i.e. the truth value of α→F is always opposite to that of α, so it can be replaced by ¬α.

The rules regarding the biconditional are as follows (they are two since the order of its constituents is irrelevant):

 biconditional: T↔α (or α↔T) is replaced by α F↔α (or α↔F) is replaced by ¬α

When α is true, α↔T is also true (a biconditional is true if and only if its constituent sentences have the same truth value). If α is false, α↔T is also false because then the truth value of the constituent sentences is different. So, α↔T always has the same truth value as α and can therefore be replaced by it. As for the second rule, when α is true, F↔α is false because then the truth value of the constituent sentences is different. When α false, F↔α is true since the truth value of the constituent sentences is the same. It turns out that F↔α always has the opposite to α truth value and can therefore be replaced by ¬α.

Here is an example of a truth-value analysis involving all logical connectives.

 [(p∧q)∨(¬p∧q)] → (q↔¬r) q: Т q: F [(p∧Т)∨(¬p∧Т)] → (Т↔¬r) [(p∧F)∨(¬p∧F)] → (F↔¬r) [p∨¬p] → ¬r [F∨F] → ¬¬r Т → ¬r F → r ¬r Т r: Т r: F F Т

This time, we started with “q” rather than “p”. We can start with whatever letter we want, but as a rule the analysis is the shortest when it starts with the letter that is most common. That is why I chose “q” here. In the first main branch (the case when “q” is true), “[p∨¬p]→¬r” is derived from “[(p∧T)∨(¬p∧T)]→(T↔¬r)” by applying the rules for T∧α (twice) and T↔α. Then “T→¬r” is derived from “[p∨¬p]→¬r” on the basis that “p∨¬p” is an obvious tautology and can therefore be replaced by “T”.

Generally, when we come to a disjunction containing a formula and its negation, such as “р∨¬р”, “qr∨¬q”, “[r→(q∧¬s)]∨¬rr”, “(pq)∨¬(pq)∨r”, etc., we can replace it with “T” since it is obviously a tautology and always has the value T. For the same reason, we can replace with “T” the conditionals and biconditionals that connect the same formula, such as “рр”, “¬(рr)→¬(рr)”, “qq”, “(qr)↔(qr)”, etc. Also, we can replace obvious contradictions with “F” – for example conjunctions containing a formula and its negation like “p∧¬p”, “q∧¬qr”, “[r→(q∨¬s)]∧¬rr”, “(pq)∧¬(pq)∧r”, etc.

Going back to the example, “T→¬r” is then replaced by “¬r” by one of the conditional rules. So far we have that when “q” is true, the whole formula has the opposite to “r” truth value. In the second branch, corresponding to the case when “q” is false, we obtain “[F∨F]→¬¬r” by applying to the previous line the rules for F∧α (twice) and F↔α. Expressions like “F∨F”, which do not include propositional letters, are replaced by “T” of “F” on the basis of the corresponding truth tables. For example, “T∧F” is replaced by “F”, “F→T” by “T”, etc. In addition, ¬¬α can be replaced by α, since the two expressions are obviously equivalent. So, substituting in “[F∨F]→¬¬r” “F” for “F∨F” and “r” for “¬¬r”, we get “F→r”, which is then replaced by “T” by one of the conditional rules. Overall, the second branch shows that when “q” is false, the whole formula is true regardless of the truth values of the other two propositional letters.

For a shorter analysis, it is advisable to begin a new branch (by replacing a propositional letter with a truth value) only after we have eliminated all occurrences of “T” or “F” in the branch we are in. For example, if we come to “(¬rq)→F”, we should first eliminate “F” in it by transforming the expression into “¬(¬rq)” using the corresponding replacement rule and only then start new branches replacing “r” or “q” with “T” and “F”.