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Proof step 12 from proof lenle (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# ay0 + # az < # ax0, unique_al(ax0), ∀ a0. a0 ∈ ax0 → a0 ∈ az ∨ a0 ∈ ay0 ⊦ ax0 = [], a ∈ ax0, ∃ ay0. (∀ a. a ∈ ax0 → a ∈ ay0) ∧ ¬ # ax0 ≤ # ay0

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);%inone: ⊦ a ∈ (b × d0)' ↔ a = b;% ⊦ # ax = 0 ↔ ax = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n + 0 = n;% ⊦ m +1 = m + 1;% ⊦ # p ' = 0 +1;% ⊦ #(ax + ax0) = # ax + # ax0;