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Proof step 7 from proof diff-twice-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ∈ y, a ∈ rmdup(y + z), ∀ y, z. (x0 \ y) \ z = x0 \ rmdup(y + z) ⊦ (x0 \ y) \ z = x0 \ rmdup(y + z)

Rule argument

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;% ⊦ a ∈ x → a ∈ x + y; (s)%rm-in-01: ⊦ a ∈ rmdup(x) → a ∈ x; (forward) from specification list-dup%a: ⊦ (x + y) + z = x + y + z; from specification list