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Proof step 24 from proof diff-union (previous step)

∀ y, z. (x0 ∪ y) \ z = (x0 \ z) ∪ (y \ z) ⊦ ((a + x0) ∪ y) \ z = ((a + x0) \ z) ∪ (y \ z)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
∀ y0, z0. rmdup(x0 + y0) \ z0 = rmdup((x0 \ z0) + (y0 \ z0)) ⊦ rmdup((a ' + x0) + y) \ z = rmdup(((a ' + x0) \ z) + (y \ z))

Rule argument

Additional information

Used simplifier rules: app: ⊦ (x + y) ∪ z = x ∪ y ∪ z;%union: ⊦ x ∪ y = rmdup(x + y);%assoc: ⊦ (x ∪ y) ∪ z = x ∪ y ∪ z;%a: ⊦ (x + y) + z = x + y + z; from specification list%rm-rm-01: ⊦ rmdup(rmdup(x) + y) = rmdup(x + y); from specification list-dup%c: ⊦ a + x = a ' + x; from specification list