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

∀ y. rev(x0 + y) = rev(y) + rev(x0) ⊦ rev((a + x0) + y) = rev(y) + rev(a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ y. rev(x0 + y) = rev(y) + rev(x0) ⊦ rev(x0 + y) = rev(y) + rev(x0)

Rule argument

Additional information

Used simplifier rules: rev-r: ⊦ rev(a ' + x) = rev(x) + a;%app-03: ⊦ x + y = z + y ↔ x = z; from specification list%Last: ⊦ x + a = x + a '; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list