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Proof step 16 from proof diff-twice-append-rev (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
∀ y, z. (x0 \ y) \ z = x0 \ (z + y) ⊦ (x0 \ y) \ z = x0 \ (z + y), a ∈ z, a ∈ y

Rule argument

Additional information

Used simplifier rules: diff-n: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y);%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list