1. | a. | φ and ψ together entail χ |
b. | ψ ⇒ φ | |
c. | φ is equivalent to itself | |
d. |
ψ is absurd
or: ψ taken by itself forms an inconsistent set |
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e. |
Γ, φ ⇒
or: Γ, φ ⇒ ⊥ (Strictly speaking, Γ, φ ⇒ ⊥ expresses entailment rather than inconsistency, but it is true if and only if φ is inconsistent with Γ.) |
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f. | Γ, ψ ⇒ φ |
2. | a. | We have supposed that Γ ⇒ φ. That is, we have supposed that φ is T in any possible world in which all members of Γ are T. But w is a world in which all members of Γ are T, so φ, too, must be T in w. |
b. | We now know that φ and all members of Γ are T in w. But we supposed that Γ, φ ⇒ ψ and we now know that all the premises of this entailment are T in w, so ψ also must be T also. | |
c. | For w to be a counterexample to Γ ⇒ ψ, it must make give ψ the value F and give all the members of Γ the value T. | |
d. | A counterexample to Γ ⇒ φ must give φ the value F and give all the members of Γ the value T. A counterexample to Γ, φ ⇒ ψ must give ψ the value F while giving φ and all the members of Γ the value T. | |
e. | We know that w gives ψ the value F and gives all the members of Γ the value T. But it also must make φ either T or F. If it does the former, it is a counterexample to Γ, φ ⇒ ψ; and if it does the latter, it is a counterexample to Γ ⇒ φ. |