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Proof step 10 from proof rmdup (previous step)

∀ 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)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 11:
∀ a0. a0 ∈ rmdup(x) → a0 = a ∨ a0 ∈ y0, ∀ x. (∀ a. a ∈ rmdup(x) → a ∈ y0) ↔ (∀ a. a ∈ x → a ∈ y0) ⊦ ∀ a0. a0 ∈ x → a0 = a ∨ a0 ∈ y0
Proof step 13:
∀ a0. a0 ∈ x → a0 = a ∨ a0 ∈ y0, ∀ x. (∀ a. a ∈ rmdup(x) → a ∈ y0) ↔ (∀ a. a ∈ x → a ∈ y0) ⊦ ∀ a0. a0 ∈ rmdup(x) → a0 = a ∨ a0 ∈ y0

Rule argument

pos: right 1

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