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Proof step 15 from proof diff-twice-append (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 16:
∀ y, z. (x0 \ y) \ z = x0 \ (y + z) ⊦ a ' + ((x0 \ y) \ z) = (a ' + x0) \ (y + z), a ∈ y, a ∈ z, a ∈ y + z

Rule argument

fma: a ∈ y + z

Additional information

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