[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 4 from proof in-vl-in-lebrefs (previous step)

∀ lref. lref ∈ q.lebrefs ↔ (∃ n. eq-entry(n, lref) ∈ q) ∧ lref ≠ none ⊦ lref ∈ q.lebrefs ∪ (eqent.lebref ≠ none ⊃ ∅ ++ eqent.lebref;∅) ↔ (∃ n. eq-entry(n, lref) = eqent ∨ eq-entry(n, lref) ∈ q) ∧ lref ≠ none

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
eqent = eq-entry(m, lref0), ∀ lref. lref ∈ q.lebrefs ↔ (∃ n. eq-entry(n, lref) ∈ q) ∧ lref ≠ none ⊦ lref ∈ q.lebrefs ∪ (lref0 ≠ none ⊃ ∅ ++ lref0;∅) ↔ (∃ n. eq-entry(n, lref) = eqent ∨ eq-entry(n, lref) ∈ q) ∧ lref ≠ none

Rule argument

spec: erasequeueentry name: elim seq: ⊦ m = eqent.pnum ∧ lref = eqent.lebref ↔ eqent = eq-entry(m, lref) subst: [eqent, m, lref] → [eqent, m, lref0]

Additional information

Used simplifier rules: ⊦ m = eqent.pnum ∧ lref = eqent.lebref ↔ eqent = eq-entry(m, lref); (elim) Used terms: eqent, m, lref0