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Proof step 15 from proof uniquelen (previous step)

a ∈ ay ↔ (∃ d, ay0, az. ay = ay0 + (a × d)' + az), unique_al(ay), unique_al(ax0), a ∈ ay, ∀ ay0. (∀ a. a ∈ ax0 ↔ a ∈ ay0) ∧ unique_al(ay0) → # ax0 = # ay0, ∀ a0. a0 = a ∨ a0 ∈ ax0 ↔ a0 ∈ ay ⊦ # ax0 + 1 = # ay, a ∈ ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
unique_al(ax0), unique_al(ay0 + (a × d)' + az), ∀ ay0. (∀ a. a ∈ ax0 ↔ a ∈ ay0) ∧ unique_al(ay0) → # ax0 = # ay0, ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az) ⊦ # ax0 = # ay0 + # az, a ∈ ax0, a ∈ ay0, a ∈ az

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);%inone: ⊦ a ∈ (b × d0)' ↔ a = b;%nuni: ⊦ unique_al(ay + (a × d)' + az) → ¬ a ∈ ay;%nuni-01: ⊦ unique_al(ay + (a × d)' + az) → ¬ a ∈ az;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%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]%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]