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Proof step 2 from proof rev-disjoint-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x; from specification list