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Proof step 3 from proof uniquelen (previous step)

unique_al([]), unique_al(ay), ∀ a. a ∈ [] ↔ a ∈ ay ⊦ # [] = # ay

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
unique_al(ay.tail), ∀ a. ¬ a ∈ ay ⊦ ay = [], ay.head.1 ∈ ay.tail

Rule argument

Additional information

Used simplifier rules: k-unique-r: ⊦ unique_al(ay) ↔ ay = [] ∨ ¬ ay.head.1 ∈ ay.tail ∧ unique_al(ay.tail);%unique-e: ⊦ unique_al([]);%empty: ⊦ ¬ a ∈ [];% ⊦ # ax = 0 ↔ ax = []; (s)% ⊦ # [] = 0; (s)