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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)