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Proof step 5 from proof diff-twice-append-rev (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
a ∈ z + y, a ∈ y, ∀ y, z. (x0 \ y) \ z = x0 \ (z + y) ⊦ (x0 \ y) \ z = (a ' + x0) \ (z + y)

Rule argument

fma: a ∈ z + y

Additional information

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