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Proof step 5 from proof lenle (previous step)

⊦ ¬ ( # ay ≤ # ax0 ∧ unique_al((a × d)' + ax0) ∧ (unique_al(ax0) ∧ (∃ ay0. (∀ a. a ∈ ax0 → a ∈ ay0) ∧ ¬ # ax0 ≤ # ay0) → ax0 = []) ∧ (∀ a0. a0 ∈ (a × d)' ∨ a0 ∈ ax0 → a0 ∈ ay))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# ay ≤ # ax0, unique_al(ax0), (∃ ay0. (∀ a. a ∈ ax0 → a ∈ ay0) ∧ ¬ # ax0 ≤ # ay0) → ax0 = [], ∀ a0. a0 ∈ ax0 ∨ a0 = a → a0 ∈ ay ⊦ a ∈ ax0

Rule argument

Additional information

Used simplifier rules: cons: ⊦ unique_al((a × d)' + ax) ↔ ¬ a ∈ ax ∧ unique_al(ax);%inone: ⊦ a ∈ (b × d0)' ↔ a = b;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;