Exportation: Some Explanations

Let me try to explain the exportation rule a few more times.

Here's the rule (I am using the triple bar to symbolize logical equivalence and the arrow to symbolize conditionals):

(p & q) -> r ≡ p -> (q -> r)

 Explanation 1: Look at the left side of the triple bar. It tells us that if p and q are true, then r is true. Look at the right side of the triple bar. It tells us that if p is true, then if q is true, then r is true. Even in plain English that sounds nearly identical. 

 Explanation 2: Look at the left side of the triple bar. It tells us, according to the meaning of conditionals, that if p is false then the whole conditional is true. Look at the right side of the triple bar. It tells us that if p is false the whole conditional is true (just remind yourself of the truth-table for p -> q).

 Explanation 3: Look at the left side of the triple bar. It tells us that if q is false the whole conditional is true. Look at the right side of the triple bar. It tells us that if q is false the whole conditional is true (again, check the truth table for conditionals to see this). 

 Explanation 4: Look at the left side of the triple bar. It tells us that if r is true the whole conditional is true. Look at the right side of the triple bar. It tells us that if r is true, the whole conditional is true. 

 Explanation 5: Look at the left side of the triple bar. It tells us that if p and q are true and r is false, then the whole conditional is false. Look at the right side of the triple bar. It tells us that if p and q are true and r is false, then the whole conditional is false. 

 Explanation 6: In propositional logic, to say that two propositions are logically equivalent is to say that each proposition implies the other one. For example, if p is logically equivalent to ~~p then p implies ~~p and ~~p implies p. So, to say that (p&q) ->r is logically equivalent to p -> (q -> r) is just to say that (p&q) ->r implies  p -> (q -> r) and  p -> (q -> r) implies (p&q) ->r. We can use conditional proof to show that. 

 Here is a conditional proof (in English) to show that (p&q) ->r implies p -> (q -> r)

 Assume (p&q) ->r for CP
Apply IMPL to (p&q) ->r
That gives us ~(p&q) v r
Apply DM ~(p&q) v r
That gives us (~p v ~q) v r
Apply ASSOC to (~p v ~q) v r
That gives us ~p v (~q v r)
Apply IMPL to ~p v (~q v r)
That gives us p -> (~q v r)
Apply IMPL to p -> (~q v r)
That gives us p -> (q -> r)
We just showed what we wanted: (p&q) ->r implies p -> (q -> r)

 Now we need to show the other direction: p -> (q -> r) implies (p&q) ->r

 I bet you can do it! I will give 2 points of extra credit to those who email me the correct proof. 

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