Is there a way to understand material implication without truth-tables?

Question

I understand how, given the truth-functional definition of the conditional, any sentence of the form

If P then Q

is truth-functionally equivalent to

Either not-P or Q

Or, to put it symbolically:  P → Q is truth-functionally equivalent to ~P v Q.

The reason why this is so is because, given the truth-tables for conditionals (P → Q), disjunctions (P v Q), and negations (~P), any sentence of the form P → Q will have the same truth-value as any sentence of the form ~P v Q.

However, suppose someone is unsure or skeptical of the truth-functional definition of the conditional. Is there some other way of showing that, for any sentence P, Q, the conditional P → Q is true if and only if the disjunction ~P v Q is true?

Answer 1

Yes. Let P and Q be any sentence(s). We will show that, given the inference rules of De Morgan’s law, Proof by Contradiction, Double Negation, Simplification/Conjunction Introduction/Elimination, Proof by Assumption, and Disjunctive Syllogism, that the conditional P → Q is true only if the disjunction ~P v Q is true, and that the disjunction ~P v Q is true only if the conditional P → Q is true.

Part 1: Suppose that

1. P → Q

For sake of argument, suppose that it’s not the case that the disjunction ~P v Q is true—i.e. suppose that

2. ~(~P v Q)

By De Morgan’s law, it follows that

3. ~~P & ~Q

By Double Negation, it follows that

4. P & ~Q

By Simplification/Conjunction Elimination it follows that

5.  P

By 1 and 5 and Modus Ponens it follows that

6. Q

By 4 and Simplification/Conjunction Elimination it follows that

7. ~Q

From 6 and 7 and Conjunction Introduction it follows that

8. Q & ~Q

But 8 is a contradiction. Therefore, by Proof by Contradiction, our assumption that ~(~P v Q) is not true—i.e. it follows that

9. ~P v Q

Therefore, given the inference rules of Proof by Contradiction, De Morgan’s law, Double Negation, Modus Ponens, and Simplification/Conjunction Introduction/Elimination, the conditional P → Q is true only if the disjunction ~P v Q is true.

Part 2: Suppose

1. ~P v Q

For sake of argument, suppose

2. P

By 2 and Double Negation it follows that

3. ~~P

By 1, 3, and Disjunctive Syllogism it follows that

4. Q

Thus, by Proof by Assumption and lines 2 through 4, if follows that if P then Q—i.e.

5. P → Q

Thus, we have shown that, given the inference rules of Proof by Assumption, Double Negation, and Disjunctive Syllogism, if the disjunction ~P v Q is true then the conditional P → Q is true.

Part 3: In Part 1 we saw that, given the inference rules of Proof by Contradiction, De Morgan’s law, Double Negation, Modus Ponens, and Simplification/Conjunction Introduction/Elimination, the conditional P → Q is true only if the disjunction ~P v Q is true. In Part 2 we saw that given the disjunction ~P v Q and the inference rules of Proof by Assumption, Double Negation, and Disjunctive Syllogism, the disjunction ~P v Q is true only if the conditional P → Q is true. What this shows is that, given these inference rules, for any sentence P, Q, the conditional P → Q is true only if the disjunction ~P v Q is true, and vice-versa—i.e. the disjunction ~P v Q is true only if the conditional P → Q is true. Therefore, so long as the rules we used in Part 1 and 2 are valid, any conditional of the form P → Q is equivalent to the disjunction ~P v Q, and this equivalence holds even without the truth-functional definition of the conditional.