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Proof step 9 from proof diff-twice-append (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 10:
a ∈ z, ∀ y, z. (x0 \ y) \ z = x0 \ (y + z) ⊦ (a ' + (x0 \ y)) \ z = (a ' + x0) \ (y + z), a ∈ y
Proof step 14:
∀ y, z. (x0 \ y) \ z = x0 \ (y + z) ⊦ a ∈ z, (a ' + (x0 \ y)) \ z = (a ' + x0) \ (y + z), a ∈ y

Rule argument

fma: a ∈ z

Additional information

Used formulas: a ∈ z