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How can we reason about “if P then Q” or “P only if Q” statements in propositional logic?

I have read in quite a few books that the proposition 'p->q' can be read as either 'if p then q' and 'p only if q'.

Let p = it rains, and q = take an umbrella.

Then with according to the first form the argument is 'If it rains, take an umbrella'. And according to the second form the argument is 'Take an umbrella only if it rains'.

Then the cases 'it rains - took an umbrella (T)', 'it did not rain - did not took an umbrella (T)', 'it rains- did not took an umbrella (F)' are trivial.

But what about the case:

did not rain - took an umbrella.

Why is this true and not for example undefined or not known?

Community
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vaweenoise
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    Note that "take an umbrella only if it rains" is a gloss for `q→p`, not `p→q`. You want "it only rains if you take an umbrella", which would certainly be true if you took an umbrella every time it rained. – Niel de Beaudrap Oct 24 '12 at 13:06
  • This reads to me like it covers very similar ground as other questions; let me know if think the problem here isn't being addressed in the other question and answer and we can maybe discuss a good reformulation of this which might address those concerns. – Joseph Weissman Oct 30 '12 at 18:30

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Also read sometimes as: p implies q. All this means is that p being true defines a value for q. When p is false, q can be anything it likes.

The "True" (not undefined or not known) in the truth table means that these values do not contradict the rules, which is true in this case. It is more that the RULE is (can be) true when p and q have these values, than saying that the rule proves these values, which is how you seem to be trying to read it.

Ryno
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  • I'm sorry. I downvoted because I read your answer incorrectly :( Apparently, if you edit the answer I can revoke my vote. – Schiphol Oct 24 '12 at 17:34
  • Mind those p's and q's! :) - edited just to see if that works. – Ryno Oct 25 '12 at 08:56