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Proof step 23 from proof in-rm-v (previous step)

eq-entry(m, lref) ∈ (lref0.vol = v ⊃ q0 - lref0.vol;eq-entry(m0, lref0) ' + (q0 - v)), ∀ m, lref, v. eq-entry(m, lref) ∈ q0 - v → (∃ eqent0. eqent0.pnum = m ∧ eqent0 ∈ q0) ⊦ lref0 = none, ∃ eqent0. eqent0.pnum = m ∧ (eqent0 = eq-entry(m0, lref0) ∨ eqent0 ∈ q0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 24:
lref0 = v0 × m1, eq-entry(m, lref) ∈ (v0 = v ⊃ q0 - v0;eq-entry(m0, lref0) ' + (q0 - v)), ∀ m, lref, v. eq-entry(m, lref) ∈ q0 - v → (∃ eqent0. eqent0.pnum = m ∧ eqent0 ∈ q0) ⊦ lref0 = none, ∃ eqent0. eqent0.pnum = m ∧ (eqent0 = eq-entry(m0, lref0) ∨ eqent0 ∈ q0)

Rule argument

spec: lebref name: elim seq: ⊦ lref ≠ none → (v = lref.vol ∧ m = lref.leb ↔ lref = v × m) subst: [lref, v, m] → [lref0, v0, m1]

Additional information

Used simplifier rules: ⊦ lref ≠ none → (v = lref.vol ∧ m = lref.leb ↔ lref = v × m); (elim) Used terms: lref0, v0, m1