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Proof step 11 from proof diff-twice-append (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 12:
a ∈ z, a ∈ y + z, ∀ y, z. (x0 \ y) \ z = x0 \ (y + z) ⊦ (x0 \ y) \ z = (a ' + x0) \ (y + z), a ∈ y

Rule argument

fma: a ∈ y + z

Additional information

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