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Proof step 1 from proof k-union-rec-y (conclusion)

⊦ a ∈ x ∨ a ∈ y → (a + x) ∪ y = x ∪ y

Heuristic simplifier applied rule simplifier

And closed the goal!

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;%rm-in-04: ⊦ a ∈ y → rmdup(a ' + x + y) = rmdup(x + y); from specification list-dup%rm-in-05: ⊦ a ∈ x → rmdup(a ' + x + y) = rmdup(x + y); from specification list-dup%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