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Proof step 2 from proof in-vl-append0 (previous step)

⊦ (∃ n0. eq-entry(n0, lref) ∈ q ∨ eq-entry(n0, lref) = eq-entry(n, lref0)) ↔ (∃ n. eq-entry(n, lref) ∈ q) ∨ lref = lref0

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
⊦ (∃ n0. eq-entry(n0, none) ∈ q ∨ eq-entry(n0, none) = eq-entry(n, lref0)) ↔ (∃ n. eq-entry(n, none) ∈ q) ∨ none = lref0
Proof step 12:
⊦ (∃ n0. eq-entry(n0, v × m) ∈ q ∨ eq-entry(n0, v × m) = eq-entry(n, lref0)) ↔ (∃ n. eq-entry(n, v × m) ∈ q) ∨ v × m = lref0

Rule argument

term: lref

Additional information

Used simplifier rules: lebref freely generated by none, ×;;