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Proof step 17 from proof diff-twice-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
∀ y, z. (x0 \ y) \ z = x0 \ rmdup(y + z) ⊦ (x0 \ y) \ z = x0 \ rmdup(y + z), a ∈ z, a ∈ y

Rule argument

Additional information

Used simplifier rules: diff-n: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y);%rm-in: ⊦ a ∈ rmdup(x + y) ↔ a ∈ x ∨ a ∈ y; from specification list-dup%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list