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Proof step 9 from proof in-rm-v (previous step)

eq-entry(m, lref) ∈ (lref0 ≠ none ∧ 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) ⊦ ∃ eqent0. eqent0.pnum = m ∧ (eqent0 = eq-entry(m0, lref0) ∨ eqent0 ∈ q0)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 10:
lref0 = none, eq-entry(m, lref) ∈ (lref0 ≠ none ∧ 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) ⊦ ∃ eqent0. eqent0.pnum = m ∧ (eqent0 = eq-entry(m0, lref0) ∨ eqent0 ∈ q0)
Proof step 22:
eq-entry(m, lref) ∈ (lref0 ≠ none ∧ 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)

Rule argument

fma: lref0 = none

Additional information

Used formulas: lref0 = none