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Proof step 12 from proof in-vl-in-lebrefs (previous step)

lref = lref0 ∨ lref ∈ q.lebrefs ⊦ #(q.lebrefs ++ lref0) = 0, lref0 = none, lref ∈ q.lebrefs, (∃ n. eq-entry(n, lref) = eq-entry(m, lref0) ∨ eq-entry(n, lref) ∈ q) ∧ lref ≠ none, (∃ n. eq-entry(n, lref) ∈ q) ∧ lref ≠ none

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
⊦ lref0 = none, lref0 ∈ q.lebrefs, ∃ n. n = m ∨ eq-entry(n, lref0) ∈ q, ∃ n. eq-entry(n, lref0) ∈ q

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)% ⊦ ¬ lref ∈ s0 → #(s0 ++ lref) = # s0 + 1; (s)% ⊦ eq-entry(m, lref) = eq-entry(m0, lref0) ↔ m = m0 ∧ lref = lref0; (ws)