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Proof Help

 
 
Guy5
 
Reply Wed 22 Nov, 2017 02:41 pm
I am in an intro logic class and need some help figuring out a couple proofs. Te first is:
1. (∃xFx & ¬Ws) → ∀x(Tx → Wx)
2. ¬∃xLx → ¬(Ft →Ws)
∴ ∀x(¬Lx) → ∀x(Tx → Wx)

The second is:
1. (Sc v Dc) → ∀xAx
. (¬Dc v Jc) → ∀xNx
∴ Ac v Nc

Any help I could get with at least getting them started is appreciated, thank you!
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carpenters
 
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Reply Fri 24 Nov, 2017 04:34 pm
@Guy5,
For the first one, assume the antecedent of the conclusion. Then use the Quantifier Shift sequent ∀x(¬Lx) → ¬∃xLx. Then using modus ponens on 2, you get ¬(Ft →Ws). The latter is the same as (Ft & ¬Ws). From Ft you can deduce ∃xFx by existential introduction. And from there using modus ponens on 1, you get the conseqent of the conclusion. Applying implication introduction is straightforward, and you get the conclusion.

For the second one, assume the negation of the conclusion in an reductio ad absurdum strategy.

So, ¬ (Ac v Nc), which is the same as (¬Ac & ¬Nc) by DeMorgan theorem. From ¬Ac, deduce ∃x¬Ax. Apply quantifier shift and you get ¬∀xAx. The same with the other conjunct, and you get ¬∀xNx. Using these two, apply modus tollens to the two premises. to get ¬(Sc v Dc) and ¬(¬Dc v Jc) respectively. Using DeMorgan you get (¬Sc & ¬Dc) and (Dc & ¬Jc). Clearly, there is a contradiction with Dc and ¬Dc. You can then with this contradiction justify the negation of the original assumption, i.e. ¬(Ac v Nc). So you then arrive at the conclusion (Ac v Nc).
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