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Proof step 15 from proof valid-eraseq-strong-notinvols (previous step)

n < # apebs, valid-eraseq-strong(apebs, vtbl, eqent + q0), valid-peb(v, m, apebs[n]), ∀ n, apebs, vtbl, v, m. n < # apebs ∧ valid-eraseq-strong(apebs, vtbl, q0) ∧ valid-peb(v, m, apebs[n]) ∧ (¬ v ∈ vtbl ∨ ¬ m < vtbl[v]) → valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) ') ∨ n ∈ q0, ¬ v ∈ vtbl ∨ ¬ m < vtbl[v] ⊦ valid-eraseq-strong(apebs, vtbl, (eqent + q0) + eq-entry(n, v × m) '), n ∈ eqent + q0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
n < # apebs, eqent.pnum < # apebs, valid-eraseq-strong(apebs, vtbl, q0), valid-peb(v, m, apebs[n]), ∀ n, apebs, vtbl, v, m. n < # apebs ∧ valid-eraseq-strong(apebs, vtbl, q0) ∧ valid-peb(v, m, apebs[n]) ∧ (¬ v ∈ vtbl ∨ ¬ m < vtbl[v]) → valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) ') ∨ n ∈ q0, ∀ v, m. (valid-peb(v, m, apebs[eqent.pnum]) ↔ eqent.lebref = v × m) ∧ (eqent.lebref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → apebs[eqent.pnum].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max), ¬ v ∈ vtbl ∨ ¬ m < vtbl[v] ⊦ eqent.pnum ∈ q0, n ∈ eqent ', n ∈ q0, ¬ eqent.pnum ∈ q0 + eq-entry(n, v × m) ' ∧ valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) ')

Rule argument

Additional information

Used simplifier rules: valid-eraseq-strong-rec: ⊦ valid-eraseq-strong(apebs, vtbl, eqent ' + q) ↔ eqent.pnum < # apebs ∧ ¬ eqent.pnum ∈ q ∧ (∀ v, m. (valid-peb(v, m, apebs[eqent.pnum]) ↔ eqent.lebref = v × m) ∧ ( eqent.lebref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → apebs[eqent.pnum].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max)) ∧ valid-eraseq-strong(apebs, vtbl, q);% ⊦ n ∈ q0 + q1 ↔ n ∈ q0 ∨ n ∈ q1; (ss)%a: ⊦ (q + q0) + q1 = q + q0 + q1; from specification list[erasequeue-base]% ⊦ eqent + q = eqent ' + q; (s)