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)) /\ (((p || q) /\ (r || s)) || ~(r || s))

⇒ logic.propositional.oroverand, 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)) /\ (p || q || ~(r || s)) /\ (r || s || ~(r || s))

⇒ logic.propositional.complor

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

⇒ logic.propositional.demorganor

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

⇒ logic.propositional.oroverand

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

⇒ logic.propositional.truezeroand

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