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Proof step 39 from proof diff-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
∀ y0, z0. rmdup(x0 + y0) \ z0 = rmdup((x0 \ z0) + (y0 \ z0)), a ∈ x0 ∨ a ∈ y, a ∈ x0 \ z ∨ a ∈ y \ z ⊦ rmdup(x0 + y) \ z = rmdup((x0 \ z) + (y \ z)), a ∈ z

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ x → rmdup(a ' + x) = rmdup(x); (s)