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Proof step 5 from proof lenle (previous step)

unique_al(ax0), # ay ≤ # ax0, (∃ ay. # ay < # ax0 ∧ unique_al(ax0.rest) ∧ (∀ a. a ∈ ax0 → a ∈ ay)) → ax0 = [] ∨ ax0.first.1 ∈ ax0.rest, ∀ a0. a0 ∈ (a × d)' ∨ a0 ∈ ax0 → a0 ∈ ay ⊦ (a × d).1 ∈ ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
unique_al(ax0), # ay ≤ # ax0, (∃ ay. # ay < # ax0 ∧ unique_al(ax0.rest) ∧ (∀ a. a ∈ ax0 → a ∈ ay)) → ax0 = [] ∨ ax0.first.1 ∈ ax0.rest, ∀ a0. a0 = a ∨ a0 ∈ ax0 → a0 ∈ ay ⊦ a ∈ ax0

Rule argument

Additional information

Used simplifier rules: inone: ⊦ a ∈ (b × d0)' ↔ a = b;%sel-02: ⊦ (a × d).1 = a; from specification pair