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Proof step 4 from proof rmdup-01 (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 5:
∀ x. (∀ a. a ∈ x → a ∈ rmdup(y0)) ↔ (∀ a. a ∈ x → a ∈ y0), ∀ a0. a0 ∈ x → a0 ∈ rmdup(a ' + y0) ⊦ ∀ a0. a0 ∈ x → a0 ∈ a ' + y0
Proof step 7:
∀ x. (∀ a. a ∈ x → a ∈ rmdup(y0)) ↔ (∀ a. a ∈ x → a ∈ y0), ∀ a0. a0 ∈ x → a0 ∈ a ' + y0 ⊦ ∀ a0. a0 ∈ x → a0 ∈ rmdup(a ' + y0)

Rule argument

pos: right 1

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