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Proof step 8 from proof rm-v-subseteq (previous step)

eqent ∈ (lref ≠ none ∧ lref.vol = v ⊃ q0 - lref.vol;eq-entry(m, lref) ' + (q0 - v)), ∀ eqent, v. eqent ∈ q0 - v → eqent ∈ q0 ⊦ eqent = eq-entry(m, lref), eqent ∈ q0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 9:
lref.vol = v, eqent ∈ q0 - lref.vol, ∀ eqent, v. eqent ∈ q0 - v → eqent ∈ q0 ⊦ eqent = eq-entry(m, lref), lref = none, eqent ∈ q0
Proof step 13:
eqent ∈ eq-entry(m, lref) ' + (q0 - v), ∀ eqent, v. eqent ∈ q0 - v → eqent ∈ q0, lref = none ∨ lref.vol ≠ v ⊦ eqent = eq-entry(m, lref), eqent ∈ q0

Rule argument

pos: left 1

Proof step 8 from proof rm-v-subseteq (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information