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Proof step 2 from proof diff-twice-append-rev (previous step)

∀ y, z. (x0 \ y) \ z = x0 \ (z + y) ⊦ ((a + x0) \ y) \ z = (a + x0) \ (z + y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ y, z. (x0 \ y) \ z = x0 \ (z + y) ⊦ ((a ' + x0) \ y) \ z = (a ' + x0) \ (z + y)

Rule argument

Additional information

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