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Proof step 2 from proof lenequal (previous step)

∀ ay, b. unique_al(ax0) ∧ unique_al(ay) ∧ # ax0 = # ay ∧ (∀ a. a ∈ ax0 → a ∈ ay) ∧ b ∈ ay → b ∈ ax0, unique_al(p + ax0), unique_al(ay), #(p + ax0) = # ay, ∀ a. a ∈ p + ax0 → a ∈ ay, b ∈ ay ⊦ b ∈ p + ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# ax0 + 1 = # ay, unique_al(ay), b ∈ ay, unique_al(p ' + ax0), ∀ ay, b. unique_al(ax0) ∧ unique_al(ay) ∧ # ax0 = # ay ∧ (∀ a. a ∈ ax0 → a ∈ ay) ∧ b ∈ ay → b ∈ ax0, ∀ a. a ∈ p ' ∨ a ∈ ax0 → a ∈ ay ⊦ b ∈ p ', b ∈ ax0

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);%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]%c: ⊦ p + ax = p ' + ax; from specification list[assoclista]