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Proof step 4 from proof lenequal (previous step)

⊦ 1 + # ax0 = # ay ∧ unique_al(ay) ∧ b ∈ ay ∧ unique_al((a × d)' + ax0) ∧ (∀ ay, b. unique_al(ax0) ∧ unique_al(ay) ∧ # ax0 = # ay ∧ (∀ a. a ∈ ax0 → a ∈ ay) ∧ b ∈ ay → b ∈ ax0) ∧ (∀ a0. a0 ∈ (a × d)' ∨ a0 ∈ ax0 → a0 ∈ ay) → b ∈ ax0 ∨ b ∈ (a × d)'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
1 + # ax0 = # ay, unique_al(ay), b ∈ ay, unique_al(ax0), ∀ ay0, b0. unique_al(ay0) ∧ # ax0 = # ay0 ∧ (∀ a. a ∈ ax0 → a ∈ ay0) ∧ b0 ∈ ay0 → b0 ∈ ax0, ∀ a0. a0 ∈ ax0 ∨ a0 = a → a0 ∈ ay ⊦ b = a, a ∈ ax0, b ∈ 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;