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Proof step 12 from proof lenle (previous step)

# ay ≤ # ax0, unique_al(ax0.rest), ∃ d, ay0, az. ay = ay0 + (a × d)' + az, ∀ a0. a0 ∈ ax0 → a0 ∈ ay ⊦ ax0 = [], a ∈ ax0, ax0.first.1 ∈ ax0.rest, ∃ ay0. # ay0 < # ax0 ∧ (∀ a. a ∈ ax0 → a ∈ ay0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# ay0 + # az < # ax0, unique_al(ax0.rest), ∀ a0. a0 ∈ ax0 → a0 ∈ ay0 ∨ a0 ∈ az ⊦ ax0 = [], ax0.first.1 ∈ ax0.rest, a ∈ ax0, ∃ ay0. # ay0 < # ax0 ∧ (∀ a. a ∈ ax0 → a ∈ ay0)

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);%inone: ⊦ a ∈ (b × d0)' ↔ a = b;%l: ⊦ # ax = 0 ↔ ax = []; from specification list-data[assoclista]%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(p ' + ax) = (# ax)+1; from specification list[assoclista]%len: ⊦ #(ax + ax0) = # ax + # ax0; from specification list[assoclista]