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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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