Conditional Proof is used to prove a conditional, an if-then statement, such as

**If an argument is sound, then it is valid.**

This post assumes that you already know how to do proofs with the rules of inference and that you’re already familiar with material equivalence and the rules of replacement. What follows in this post is based on this video, which goes over some examples of conditional proof, then assigns some proofs for you to do on your own. If you would like to try the examples before seeing the solutions, just page down before watching the video. Solutions to the homework, and the video going over them, are at the bottom of this post.

To use conditional proof to prove a conditional, start by assuming its antecedent. Using its antecedent as a premise, construct a proof for its consequent. From the assumption of the antecedent to the conclusion of the consequent, each line of your argument should occur in a new scope. This is like scoping in programming languages. Here’s an example from PHP:

function set_a ($val) { local $a; $a = $val; } $a = 6; set_a(5); echo $a;

In this brief example, $a is used as both a global variable and as a local variable. If you don’t know any better, you might expect the last line to print 5, but it will actually print 6. This is because the variable $a in the function was at a higher scope than $a in the main part of the program. When the set_a function ended, its scope disappeared from the program, and whatever was set in that scope was no longer available. It will be like this in symbolic logic proofs too. When a scope closes, you can no longer use any lines from that scope in your proof. This restriction is to keep you from drawing illogical conclusions. With this restriction in place, conditional proof lets you make assumptions to prove conditionals without letting your prove anything you assume. Here is a brief summary of conditional proof:

To prove P ⊃ Q Assume P Derive Q Conclude P ⊃ Q

In this first example, notice that I use the | to set off a new scope. When I close the scope, I cease to place the | at the beginning of the line. Once the scope is closed, the rule of conditional proof can be used on the whole scope to conclude a conditional whose antecedent is the assumption that started the new scope and whose consequent is the last line of the scope. You may not conclude anything from individual lines of a closed scope.

1. P ⊃ Q // Premise Prove: P ⊃ (Q ∨ R) 2. | P // Assumption 3. | Q // 1,2 Modus Ponens 4. | Q ∨ R // 3 Addition 5. P ⊃ (Q ∨ R) // 2-4 Conditional Proof

From here on out, I will use a spoiler shortcode to hide the solutions in case you would like to try them before looking at them. If you need any help understanding the solutions, watch the video.

1. P ⊃ Q // Premise 2. (Q ∨ $) ⊃ (S & T) // Premise Prove: P ⊃ S

1. P ⊃ Q // Premise 2. R ⊃ S // Premise 3. T ⊃ U // Premise 4. P ∨ (R ∨ T) // Premise Prove: Q ∨ (S ∨ U)

Prove: P ⊃ (P ∨ P)

Without using Tautology, Prove: P ≡ (P & P)

Without using Tautology, Prove: P ≡ (P ∨ P)

## Homework

These proofs were assigned as homework at the end of the video. Solutions are hidden by spoiler shortcodes, and the video going over the solutions is embedded below the proofs.

Without using Commutation, Prove: (P & Q) ≡ (Q & P)

Without using Transposition, Prove: (P ⊃ Q) ≡ (~Q ⊃ ~P)

Without using Association, Prove: (P & (Q & R)) ≡ ((P & Q) & R)

1. P ⊃ Q // Premise 2. R ⊃ ~Q // Premise Prove: ~(P & R)

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