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Proof step 13 from proof uniquelen (previous step)

a = a ∨ a ∈ ax0 ↔ a ∈ ay, unique_al(ax0), unique_al(ay), ∀ a0. a0 = a ∨ a0 ∈ ax0 ↔ a0 ∈ ay, ∀ ay0. unique_al(ay0) ∧ (∀ a. a ∈ ax0 ↔ a ∈ ay0) → # ax0 = # ay0 ⊦ 1 + # ax0 = # ay, a ∈ ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
unique_al(ay), unique_al(ax0), a ∈ ay, ∀ ay0. unique_al(ay0) ∧ (∀ a. a ∈ ax0 ↔ a ∈ ay0) → # ax0 = # ay0, ∀ a0. a0 = a ∨ a0 ∈ ax0 ↔ a0 ∈ ay ⊦ 1 + # ax0 = # ay, a ∈ ax0

Proof step 13 from proof uniquelen (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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