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Proof step 22 from proof diff-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: diff-in: ⊦ a ∈ x \ y ↔ a ∈ x ∧ ¬ a ∈ y;%a: ⊦ (x + y) + z = x + y + z; from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%c: ⊦ a + x = a ' + x; from specification list