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Thread: Modal sequents

  1. #1
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    Modal sequents

    1. A sequent with only atoms is valid iff some formula occurs on
    both sides.
    2. A, ϕ ⇒ B iff A ⇒ B, ϕ
    3. A ⇒ B, ϕ iff A, ϕ ⇒ B
    4. A, ϕ ∧ ψ ⇒ B iff A, ϕ, ψ ⇒ B
    5. A ⇒ B, ϕ ∧ ψ iff both A ⇒ B, ϕ and A ⇒ B, ψ

    These are facts about modal sequents...
    How do we prove them?

    For example 2. A, ϕ ⇒ B iff A ⇒ B, ϕ
    Where A and B = finite sets of wwfs
    We suppose A, ϕ ⇒ B first. Can we prove it without using contradiction method?
    Supposing A not ⇒ B, ϕ and showing contradiction.
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  2. #2
    MHF Contributor
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    Welcome to the forum.

    A, ϕ ⇒ B iff A ⇒ B, ϕ
    The sequent A, ϕ ⇒ B is true, by definition, if $\displaystyle \bigwedge A\land\neg\phi\to\bigvee B$ is true and A ⇒ B, ϕ is true iff $\displaystyle \bigwedge A\to\bigvee B\lor\phi$ is true. Here $\displaystyle \bigwedge A$ denotes the conjunction of all formulas in A, and $\displaystyle \bigvee B$ denotes the disjunction of formulas in B. In the following, however, let's pretend that A is $\displaystyle \bigwedge A$ and B is $\displaystyle \bigvee B$.

    So, assume that $\displaystyle A\land\neg\phi\to B$ is true; we need to show $\displaystyle A\to B\lor\phi$. Assume A. If $\displaystyle \phi$ is true, then the conclusion $\displaystyle B\lor\phi$ is true and we are done. If $\displaystyle \neg\phi$ is true, then the first assumption implies B, so again the conclusion $\displaystyle B\lor\phi$ is true.

    For the converse, assume $\displaystyle A\to B\lor\phi$ is true; we need to show that $\displaystyle A\land\neg\phi\to B$ is true. Assume A and $\displaystyle \neg\phi$. Then the first assumption implies $\displaystyle B\lor\phi$. However, since $\displaystyle \neg\phi$ is true, B is also true, as required.

    Ultimately, each of the equivalences in your question corresponds to a tautology, which can be verified using a truth table. For example, (5) corresponds to $\displaystyle (A\to B\lor (\phi\land\psi))\leftrightarrow (A\to B\lor\phi)\land (A\to B\lor\psi)$.

    Why do you refer to these sequents as "modal"? They seem to come from regular, not modal logic.
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