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Proof step 10 from proof lenequal (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# ax0 = # ay0 + # az, unique_al(ay0 + (a × d)' + az), unique_al(ax0), ∀ ay0, b0. # ax0 = # ay0 ∧ unique_al(ay0) ∧ b0 ∈ ay0 ∧ (∀ a. a ∈ ax0 → a ∈ ay0) → b0 ∈ ax0, ∀ a0. a0 ∈ ax0 → a0 ∈ ay0 ∨ a0 ∈ az, b ∈ ay0 ∨ b ∈ az ⊦ b = a, b ∈ ax0, 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-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclist]% ⊦ # p ' = 0 +1; (s)% ⊦ #(ax + ax0) = # ax + # ax0; (s)