Exercise logic.propositional.consequence

Description
Prove that formula is a logical consequence of a set of formulas

Derivation

Final term is not finished

p <-> r, q <-> s => ((p -> q) /\ (r -> s)) || (~(p -> q) /\ ~(r -> s))

⇒ logic.propositional.defimpl, initial=TList [TCon logic1.equivalent [TVar "p",TVar "r"],TCon logic1.equivalent [TVar "q",TVar "s"]]

p <-> r, q <-> s => ((~p || q) /\ (r -> s)) || (~(p -> q) /\ ~(r -> s))

⇒ logic.propositional.defimpl

p <-> r, q <-> s => ((~p || q) /\ (r -> s)) || (~(~p || q) /\ ~(r -> s))

⇒ logic.propositional.defimpl

p <-> r, q <-> s => ((~p || q) /\ (r -> s)) || (~(~p || q) /\ ~(~r || s))

⇒ logic.propositional.demorganor

p <-> r, q <-> s => ((~p || q) /\ (r -> s)) || (~~p /\ ~q /\ ~(~r || s))

⇒ logic.propositional.demorganor

p <-> r, q <-> s => ((~p || q) /\ (r -> s)) || (~~p /\ ~q /\ ~~r /\ ~s)

⇒ logic.propositional.notnot

p <-> r, q <-> s => ((~p || q) /\ (r -> s)) || (p /\ ~q /\ ~~r /\ ~s)

⇒ logic.propositional.notnot

p <-> r, q <-> s => ((~p || q) /\ (r -> s)) || (p /\ ~q /\ r /\ ~s)

⇒ fakeabsorption

p <-> r, q <-> s => ((~p || q) /\ (r -> s)) || (p /\ ~q /\ r /\ ~s)