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Proof step 3 from proof diff-twice-append (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a ∈ y, ∀ y, z. (x0 \ y) \ z = x0 \ (y + z) ⊦ ((a ' + x0) \ y) \ z = (a ' + x0) \ (y + z)
Proof step 8:
∀ y, z. (x0 \ y) \ z = x0 \ (y + z) ⊦ a ∈ y, ((a ' + x0) \ y) \ z = (a ' + x0) \ (y + z)

Rule argument

fma: a ∈ y

Additional information

Used formulas: a ∈ y