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

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a ∈ z, ∀ y, z. (x0 + y) \ z = (x0 \ z) + (y \ z) ⊦ ((a ' + x0) + y) \ z = ((a ' + x0) \ z) + (y \ z)
Proof step 6:
∀ y, z. (x0 + y) \ z = (x0 \ z) + (y \ z) ⊦ a ∈ z, ((a ' + x0) + y) \ z = ((a ' + x0) \ z) + (y \ z)

Rule argument

fma: a ∈ z

Additional information

Used formulas: a ∈ z