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Proof step 10 from proof uniquelen (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
unique_al(ay), unique_al(ax0), ∀ ay0. unique_al(ay0) ∧ (∀ a. a ∈ ax0 ↔ a ∈ ay0) → # ax0 = # ay0, ∀ a0. a0 ∈ ax0 ∨ a0 = a ↔ a0 ∈ ay ⊦ 1 + # ax0 = # 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;