Level 5.1: Contradiction & Consistency (P, Q)
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In previous courses we used truth-tables to identify logical relationships within pairs of logical formulae, and to test argument forms (otherwise known as “sequents”) for validity. Truth-tables are very useful tools. However, as we saw in the last course, truth-tables can become unwieldy when we are dealing with more than two simple sentences.
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When we only have two simple sentences, our truth-table are only four lines deep, but with each additional simple sentence the number of lines doubles. Thus, if we have four simple sentences our truth-table will be thirty-two lines deep! And with thirty two lines to get right, there is a significant chance we will make a mistake. This is where the indirect method comes in handy.
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In this lesson we will use the indirect method to test pairs of formulae for consistency and contradiction. We will start with pairs that contain only two simple sentences. But in the second half of this course, we will move onto examples involving three simples sentences.
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The general approach to using the indirect method to test for Consistency & Contradiction is as follows. First, we assume that both formulae in the pair are true. Next, we check if this assumption leads us into contradiction. If it does, then we know that it is impossible for both formulae to be true at the same time: i.e. that our pair is contradictory. However, if our assumption does not lead us into contradiction, then we know it is possible for both formulae to be true at the same time: i.e. that our pair is consistent.
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Example 1
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Let’s begin. First we assign T to each of the two formulae in the pair, then carry any simple-sentence truth-values that have already been assigned to all other instances of that simple sentence:
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| Q | ((P | → | Q) | x∨ | Q) | ||
| T | T | T | T |
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Second, we must use TABLE 1 to assign a numerical value to each of our two formulae:
Table 1 – Numerical Values
| Simple Sentence | 1 |
| Negated Simple Sentence | 1 |
| Conjunction | 1 |
| Negated Conjunction | 3 |
| Inclusive Disjunction | 3 |
| Negated Inclusive Disjunction | 1 |
| Exclusive Disjunction | 2 |
| Negated Exclusive Disjunction | 2 |
| Conditional | 3 |
| Negated Conditional | 1 |
| Bi-Conditional | 2 |
| Negated Bi-Conditional | 2 |
Here’s how this looks when applied to our example:
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| 1 | 2 | ||||||
| Q | ((P | → | Q) | x∨ | Q) | ||
| T | T | T | T |
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The formulae with the lowest numerical value is our ANCHOR FORMULA: in this case, our left-hand formula “Q”.
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The third step is to consistently complete the remaining truth-values for the ANCHOR FORMULA then carry any newly-assigned simple-sentences truth-values across to any other instances of the same simple-sentence. (Note: if, as in our example, the formula with the lowest value is a simple sentence then this step is already complete):
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| 1 | 2 | ||||||
| Q | ((P | → | Q) | x∨ | Q) | ||
| T | T | T | T |
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The fourth step is to complete the truth-values for the NON-ANCHOR FORMULA. If the pair of formulae is consistent you will be able to do this consistently. However, if the pair is contradictory this will not be possible. Here’s how this will play out for our example.
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As Q, the consequent of the conditional “(P → Q)”, is true, the entire conditional will be true regardless of the truth-value of the antecedent of that conditional, “P”.
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| 1 | 2 | ||||||
| Q | ((P | → | Q) | x∨ | Q) | ||
| T | T | T | T | T |
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However, we still need to fill out the truth-value for “P”. We do this by starting with T (and seeing if this yields a consistent truth-value assignment for the entire formula)
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| 1 | 2 | ||||||
| Q | ((P | → | Q) | x∨ | Q) | ||
| T | T | T | T | T | T |
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As this assignment is contradictory (the overall truth-value of an exclusive disjunction with two truth disjuncts should be F and not T) we try the other possible truth-value assignment for “P”, i.e. F, and check again to see if the truth-value assignment for the entire formula is consistent:
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| 1 | 2 | ||||||
| Q | ((P | → | Q) | x∨ | Q) | ||
| T | F | T | T | T | T |
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If the truth-value of the consequent of a conditional is T, then changing the truth-value of the antecedent of that conditional from T to F will not change the truth-value of the entire conditional. Thus we are still left with an exclusive conditional with two true disjuncts. And the truth-value of an exclusive disjunction with two true disjuncts should be F, rather than T. Thus, we still have a contradiction in the NON-ANCHOR FORMULA (i.e. the right-hand formula). And thus, we can now conclude that our pair is contradictory.
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Indirect Method Procedure: Consistency & Contradiction
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Here’s the procedure we just applied:
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(1) Assign T to the main operator of each formula (this move constitutes our assumption that both formulae are true) then carry any simple-sentence truth-values to other instances of that same simple sentence.
| Simple Sentence | 1 |
| Negated Simple Sentence | 1 |
| Conjunction | 1 |
| Negated Conjunction | 3 |
| Inclusive Disjunction | 3 |
| Negated Inclusive Disjunction | 1 |
| Exclusive Disjunction | 2 |
| Negated Exclusive Disjunction | 2 |
| Conditional | 3 |
| Negated Conditional | 1 |
| Biconditional | 2 |
| Negated Biconditional | 2 |
(2) Assign a numerical value to each formulae using the Numerical Value Table. The formula with the lowest numerical value is our ANCHOR FORMULA (If the two formulae have the same numerical value, then the formula on the left is the ANCHOR FORMULA)
(3) Consistently complete the truth-values for the ANCHOR FORMULA (If there is more than one simple-sentence truth-value assignment that yields a consistent overall truth-value assignment for the ANCHOR FORMULA, start with the one in the highest position in TABLE 2) then carry any newly assigned simple-sentence truth-values to other instances of that same simple sentence.
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Table 2 – Truth-Value Table
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| P | Q | ||||
| T | T | ||||
| T | F | ||||
| F | T | ||||
| F | F |
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(4) If possible, consistently complete the truth-values for the NON-ANCHOR FORMULA. If there is more than one way to do this, use the consistent simple-sentence truth-value assignment that is highest on the following table:
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| P | Q | ||||
| T | T | ||||
| T | F | ||||
| F | T | ||||
| F | F |
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If this is not possible, return to the anchor formula and try the next highest consistent truth-value assignment and return to step (4). If this still fails to yield a consistent overall truth-value assignment, your pair is contradictory.
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Example 2
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Let’s work through another example. Step one is to assign the truth-value T to each of the two formulae:
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| ~ | (P | → | Q) | ~ | Q | ||
| T | T |
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Step 2 is to assign a numerical value to each formulae:
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| 1 | 1 | ||||||
| ~ | (P | → | Q) | ~ | Q | ||
| T | T |
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As both formulae in the pair have a score of 1, the lefthand formula “~(P → Q)” becomes our ANCHOR FORMULA.
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The third step is to carry any simple-sentence truth-values that have already been assigned to all other instances of that simple sentence. This step is not relevant to us this time around, as we have not yet assigned truth-values to any simple sentence.
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Step 3 is to consistently fill out the truth-values for our ANCHOR FORMULA, then carry any simple-sentence truth-values over to other instances of the same simple sentence:
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| 1 | 1 | ||||||
| ~ | (P | → | Q) | ~ | Q | ||
| T | T | F | F | T | F | – |
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Step 4 is (if possible) to consistently complete the truth-values for the formula with the lower numerical value. In this case, this is possible – and in fact has already been done – so we can conclude that our pair of formulae is consistent!