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To show that P → ¬P ∈ Γ only if P → Q ∈ Γ, would I have to use soundness, completeness, or could I prove it using derivation rules?

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    I assume that you are mening maximal consistent... – Mauro ALLEGRANZA Nov 4 '16 at 7:14
  • No, I think he wants a formal proof of the 'principle of explosion' That you can derive anything from a single contradiction. Is that a reasonable interpretation? – user9166 Nov 4 '16 at 18:02
  • @jobermark I'm not quite sure. This is just a question on my assignment but we only went over things such as if a set is maximal then it is consistent IFF it is satisfiable. But we did skim over contradictions. – K.Wong Nov 4 '16 at 18:24
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If Γ is maximally consistent set, then there is a unique valuation v such that

v(ψ) = 1 for all ψ ∈ Γ.

Thus, if P → ¬P ∈ Γ, we must have v(P → ¬P) = 1.

This means : v(P)=0 and thus : v(P → Q)=1 i.e. P → Q ∈ Γ.

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    The problem has "only if", so you have to assume P → ¬P ∈ Γ, not the other way around. – Eliran Nov 4 '16 at 15:27
  • @EliranH I thought that if "only if" was between two sentences such as "P only if Q," then it would be "If Q, then P"? – K.Wong Nov 4 '16 at 17:55
  • @K.Wong No, "P only if Q" is equivalent to "if P then Q". For example: "It's snowing only if it's cold" <=> "If it's snowing, then it's cold." – Eliran Nov 4 '16 at 19:07
  • @EliranH Ah okay. What I have now is that.. If P → ¬P ∈ Γ, then (P → ¬P) = 1. So, P = 0. Case 1: P = 0, and Q can be either 0 or 1 since it won't affect the sentence being T. So, P = 0 and Q = 0 or 1, and this (P → Q) = 1. Is this correct? – K.Wong Nov 4 '16 at 19:37

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