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Proof step 33 from proof diff-union (previous step)

¬(a ∈ x0 ∨ a ∈ y), a ∈ z, ∀ y0, z0. rmdup(x0 + y0) \ z0 = rmdup((x0 \ z0) + (y0 \ z0)) ⊦ (a ' + rmdup(x0 + y)) \ z = rmdup((x0 \ z) + (y \ z))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
a ∈ z, ∀ y0, z0. rmdup(x0 + y0) \ z0 = rmdup((x0 \ z0) + (y0 \ z0)) ⊦ rmdup(x0 + y) \ z = rmdup((x0 \ z) + (y \ z)), a ∈ x0, a ∈ y

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