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Proof step 35 from proof diff-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
∀ y0, z0. rmdup(x0 + y0) \ z0 = rmdup((x0 \ z0) + (y0 \ z0)) ⊦ (a ∈ x0 ∨ a ∈ y ⊃ rmdup(x0 + y);a ' + rmdup(x0 + y)) \ z = rmdup((a ' + (x0 \ z)) + (y \ z)), a ∈ z

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list