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Proof step 17 from proof diff-union (previous step)

¬ ¬ a ∈ z, ∀ z. rmdup(y0) \ z = rmdup(y0 \ z) ⊦ (a ' + rmdup(y0)) \ z = rmdup((a ' + y0) \ z), a ∈ y0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;%a: ⊦ (x + y) + z = x + y + z; from specification list