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Proof step 12 from proof diff-twice-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
a ∈ z, ∀ y, z. (x0 \ y) \ z = x0 \ (y + z) ⊦ (x0 \ y) \ z = x0 \ (y + z), a ∈ y

Rule argument

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;%in-12: ⊦ a ∈ y → a ∈ x + y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list