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Proof step 5 from proof uniquelen (previous step)

unique_al(ay.rest), ∀ a. ¬ a ∈ ay ⊦ ay = [], ay.first.1 ∈ ay.rest, ay.first.1 ∈ ay

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
ay = p + ax0, unique_al(ax0), ∀ a. ¬ a ∈ ay ⊦ ay = [], p.1 ∈ ax0, p.1 ∈ ay

Rule argument

spec: list-data inst: assoclista name: elim-list-c seq: ax ≠ [] ⊦ p = ax.first ∧ ax0 = ax.rest ↔ ax = p + ax0 subst: [ax, p, ax0] → [ay, p, ax] flags:

Additional information

Used simplifier rules: ax ≠ [] ⊦ p = ax.first ∧ ax0 = ax.rest ↔ ax = p + ax0; (elim) Used terms: ay, p, ax