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Proof step 3 from proof lenequal (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
1 + # ax0 = # ay, unique_al(ay), unique_al(p ' + ax0), b ∈ ay, ∀ ay, b. # ax0 = # ay ∧ unique_al(ay) ∧ unique_al(ax0) ∧ b ∈ ay ∧ (∀ a. a ∈ ax0 → a ∈ 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[assoclist]% ⊦ # p ' = 0 +1; (s)% ⊦ #(ax + ax0) = # ax + # ax0; (s)% ⊦ p + ax = p ' + ax; (s)