# Simple logic question

Can anyone tell me if the universal quantifier distributes over material conditionals in standard first order logic? Does (v)(Mv → Pv) imply {v)Mv -> (v)Pv? Does the implication work the other way too? Thanks.

HallsofIvy
Homework Helper
If I understand your notation correctly your first sentence says "for all v, if Mv is true then Pv is true" while your second is "if there exist v such that Mv is true then Pv is true for all v". No, they are not at all the same. In particular the "v" in Pv is not the same as the "v" in Mv and it really should be {v}Mv->(u)Pu. That is "if Mv is true for some v, then Pu is true for all u.

For example, if M is "gets an A on every tests" and P is "gets an A for the course", your first statement is "If any student in the class gets an A on every test, then he/she will get an A for the course" while the second statement says "If any student in the class gets an A on every test, then every student in the class will get an A for the course". Not at all the same!

I'm not sure what you meant by "does the implication work the other way". Certainly M-> P is NOT the same as P->M. "If it rains today, I will take my umbrella" does not imply "if I take my umbrella it will rain today".

The first v in the second sentence is supposed to be universally quantified like in the first sentence. It looks like I made a typo and put in some kind of bracket instead of a left parenthesis. You read the first one right, but the second is supposed to be "if for all v, Mv is true, then for all v, Pv is true". All of the quantifiers are universal, none are existential.

Let me try to explain better. If (v)(Mv → Pv) is true, can you infer (v)Mv → (v)Pv? If (v)Mv → (v)Pv is true, can you infer (v)(Mv → Pv)?

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By the way I think its valid to infer (v)Mv → (v)Pv if (v)(Mv → Pv) is true. I think that it is not valid to infer (v)(Mv → Pv) from (v)Mv → (v)Pv. I would still like somebody to confirm for sure. I am more confident in the one I think is valid than the other. Both of the inferences could be valid.

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A counterexample for the second could be the two-element set {a, b}, where Ma = true, Pa = false (so that Ma -> Pa is false, thus (v)(Mv -> Pv) is false), and Mb = false (so that (v)Mv is false, and therefore (v)Mv -> anything is true).

A counterexample for the second could be the two-element set {a, b}, where Ma = true, Pa = false (so that Ma -> Pa is false, thus (v)(Mv -> Pv) is false), and Mb = false (so that (v)Mv is false, and therefore (v)Mv -> anything is true).
Yup, your counterexample works. Thanks. Do you agree that the first is valid?

For any set, let S be the subset of v's for which Mv is true. Now, since (v)(Mv -> Pv), for each element in S you can apply 'modus ponens' and get Pv true for all v in S. In particular, if (v)Mv is true then S is the whole set.

P.S.: That is to say, "yes". :)

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