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Proof step 9 from proof lenequal (previous step)

1 + # ax0 = # ay, unique_al(ay), b ∈ ay, unique_al(ax0), ∃ d, ay0, az. ay = ay0 + (a × d)' + az, ∀ a0. a0 ∈ ax0 → a0 ∈ ay, ∀ ay0, b0. unique_al(ay0) ∧ # ax0 = # ay0 ∧ (∀ a. a ∈ ax0 → a ∈ ay0) ∧ b0 ∈ ay0 → b0 ∈ ax0 ⊦ b = a, a ∈ ax0, b ∈ ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# ax0 = # ay0 + # az, unique_al(ay0 + (a × d)' + az), unique_al(ax0), ∀ ay0, b0. unique_al(ay0) ∧ # ax0 = # ay0 ∧ (∀ a. a ∈ ax0 → a ∈ ay0) ∧ b0 ∈ ay0 → b0 ∈ ax0, ∀ a0. a0 ∈ ax0 → a0 ∈ az ∨ a0 ∈ ay0 ⊦ b = a, a ∈ ax0, b ∈ ax0, a ∈ ay0, a ∈ az, ¬ b ∈ az ∧ ¬ b ∈ ay0

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;% ⊦ m + n = m + n0 ↔ n = n0;% ⊦ (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;