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The following steps lead you to construct a proof of the law for lemmas
if Γ,φ ⇒ ψ and Γ ⇒ φ, then Γ ⇒ ψ
Begin by supposing that Γ, φ ⇒ ψ and Γ ⇒ φ are both true. We want to show that, under this supposition, Γ ⇒ ψ is also true. To do that, we consider any possible world w in which all members of Γ are true and try to show that ψ is true in w.
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a.
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Our supposition that Γ, φ ⇒ ψ and Γ ⇒ φ are both true combined with what we know about w enables us to conclude that φ is true. Why?
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b. |
Adding the information that φ is true in Γ to what we already knew, we can conclude that ψ is true. Why?
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So, knowing about w only that all members of Γ were true, we are able to conclude that ψ is true. And that shows us that ψ is true in every world in which all members of Γ are true, which means that Γ ⇒ ψ.
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Another approach to proving the law is to show that Γ ⇒ ψ fails only if at least one of Γ, φ ⇒ ψ and Γ ⇒ φ fails. The following three steps show this:
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c. |
Suppose that w is a counterexample to Γ ⇒ ψ. What truth values do ψ and the members of Γ have in w?
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What truth values are needed to have a counterexample to Γ ⇒ φ? To have a counterexample to Γ, φ ⇒ ψ?
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e. |
The world w from c will be a counterexample to either Γ, φ ⇒ ψ or Γ ⇒ φ. Why?
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