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Proof step 16 from proof in-rm-v-same (previous step)

eq-entry(n, v × m) ∈ (v0 = v ⊃ q0 - v;eq-entry(m0, v0 × m1) ' + (q0 - v)), ∀ n, v, m. ¬ eq-entry(n, v × m) ∈ q0 - v ⊦ v0 = v

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
eq-entry(n, v × m) ∈ q0 - v, ∀ n, v, m. ¬ eq-entry(n, v × m) ∈ q0 - v ⊦ eq-entry(n, v × m) = eq-entry(m0, v0 × m1), v0 = v

Rule argument

Additional information

Used simplifier rules: ⊦ v × m = v0 × m0 ↔ v = v0 ∧ m = m0; (ws)% ⊦ eq-entry(m, lref) = eq-entry(m0, lref0) ↔ m = m0 ∧ lref = lref0; (ws)%a: ⊦ (q + q0) + q1 = q + q0 + q1; from specification list[erasequeue-base]% ⊦ eqent ∈ eqent0 ' + q ↔ eqent = eqent0 ∨ eqent ∈ q; (s)