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Proof step 6 from proof diff-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ y, z. (x0 + y) \ z = (x0 \ z) + (y \ z) ⊦ (x0 + y) \ z = (x0 \ z) + (y \ z), a ∈ z

Rule argument

Additional information

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