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Proof step 9 from proof rmdup (previous step)

∀ x. (∀ a. a ∈ rmdup(x) → a ∈ y0) ↔ (∀ a. a ∈ x → a ∈ y0) ⊦ (∀ a0. a0 ∈ rmdup(x) → a0 ∈ a + y0) ↔ (∀ a0. a0 ∈ x → a0 ∈ a + y0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ x. (∀ a. a ∈ rmdup(x) → a ∈ y0) ↔ (∀ a. a ∈ x → a ∈ y0) ⊦ (∀ a0. a0 ∈ rmdup(x) → a0 = a ∨ a0 ∈ y0) ↔ (∀ a0. a0 ∈ x → a0 = a ∨ a0 ∈ y0)

Rule argument

Additional information

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