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Proof step 9 from proof firstsub (previous step)

a ∈ x0 + y, ∀ a0. a0 ∈ a ' + x0 → a0 ∈ y, ∀ y. (∀ a. a ∈ x0 → a ∈ y) → rmdup(x0 + y) = rmdup(y) ⊦ rmdup(x0 + y) = rmdup(y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ y. (∀ a. a ∈ x0 → a ∈ y) → rmdup(x0 + y) = rmdup(y), ∀ a0. a0 ∈ a ' ∨ a0 ∈ x0 → a0 ∈ y, a ∈ x0 ∨ a ∈ y ⊦ rmdup(x0 + y) = rmdup(y)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)