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Proof step 35 from proof in-rm-v (previous step)

∀ m, lref, v. eq-entry(m, lref) ∈ q0 - v → (∃ eqent0. eqent0.pnum = m ∧ eqent0 ∈ q0), eq-entry(m, lref) = eq-entry(m0, v0 × m1) ∨ eq-entry(m, lref) ∈ q0 - v ⊦ v0 = v, ∃ eqent0. eqent0.pnum = m ∧ (eqent0 = eq-entry(m0, v0 × m1) ∨ eqent0 ∈ q0)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 36:
eq-entry(m, lref) ∈ q0 - v, ∀ m, lref, v. eq-entry(m, lref) ∈ q0 - v → (∃ eqent0. eqent0.pnum = m ∧ eqent0 ∈ q0), eq-entry(m, lref) = eq-entry(m0, v0 × m1) ∨ eq-entry(m, lref) ∈ q0 - v ⊦ v0 = v, ∃ eqent0. eqent0.pnum = m ∧ (eqent0 = eq-entry(m0, v0 × m1) ∨ eqent0 ∈ q0)
Proof step 44:
∀ m, lref, v. eq-entry(m, lref) ∈ q0 - v → (∃ eqent0. eqent0.pnum = m ∧ eqent0 ∈ q0), eq-entry(m, lref) = eq-entry(m0, v0 × m1) ∨ eq-entry(m, lref) ∈ q0 - v ⊦ v0 = v, eq-entry(m, lref) ∈ q0 - v, ∃ eqent0. eqent0.pnum = m ∧ (eqent0 = eq-entry(m0, v0 × m1) ∨ eqent0 ∈ q0)

Rule argument

fma: eq-entry(m, lref) ∈ q0 - v

Additional information

Used formulas: eq-entry(m, lref) ∈ q0 - v