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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;%in-11: ⊦ a ∈ x → a ∈ x + y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list