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Proof step 5 from proof diff-append (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: y, z

Additional information

Used terms: y, z