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Proof step 17 from proof diff-twice-append (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: y, z

Additional information

Used terms: y, z