# Modal sequents

• Feb 3rd 2011, 05:32 PM
KlausAir
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.
• Feb 4th 2011, 12:26 AM
emakarov
Welcome to the forum.

Quote:

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

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

For the converse, assume $A\to B\lor\phi$ is true; we need to show that $A\land\neg\phi\to B$ is true. Assume A and $\neg\phi$. Then the first assumption implies $B\lor\phi$. However, since $\neg\phi$ 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 $(A\to B\lor (\phi\land\psi))\leftrightarrow (A\to B\lor\phi)\land (A\to B\lor\psi)$.

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