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Proof step 2 from proof rev-disjoint (previous step)

∀ x. disj(x, rev(y0)) ↔ disj(x, y0) ⊦ disj(x, rev(a + y0)) ↔ disj(x, a + y0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ x. disj(x, rev(y0)) ↔ disj(x, y0) ⊦ disj(x, rev(y0) + a ') ↔ disj(x, a ' + y0)

Rule argument

Additional information

Used simplifier rules: rev-app: ⊦ rev(x + y) = rev(y) + rev(x);%rev-one: ⊦ rev(a ') = a ';%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)