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Proof step 8 from proof diff-twice-append-rev (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ y, z. (x0 \ y) \ z = x0 \ (z + y) ⊦ (a ' + (x0 \ y)) \ z = (a ' + x0) \ (z + y), a ∈ y

Rule argument

Additional information

Used simplifier rules: diff-n: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y);%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list