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Proof step 23 from proof in-vl-in-lebrefs (previous step)

eq-entry(n, lref) ∈ q, ∀ lref. lref ∈ q.lebrefs ↔ (∃ n. eq-entry(n, lref) ∈ q) ∧ lref ≠ none ⊦ eq-entry(n, lref) = eq-entry(m, lref0), lref = lref0, lref = none, lref0 = none, lref ∈ q.lebrefs

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 24:
none ∈ q.lebrefs ↔ (∃ n. eq-entry(n, none) ∈ q) ∧ none ≠ none, eq-entry(n, lref) ∈ q, ∀ lref. lref ∈ q.lebrefs ↔ (∃ n. eq-entry(n, lref) ∈ q) ∧ lref ≠ none ⊦ eq-entry(n, lref) = eq-entry(m, lref0), lref = none, lref = lref0, lref0 = none, lref ∈ q.lebrefs

Rule argument

pos: left 2 terms: none

Additional information

Used terms: none