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Proof step 4 from proof valid-eraseq-strong-imp-q (previous step)

valid-eraseq-strong(apebs, vtbl, eqent + q0), ∀ apebs, vtbl. valid-eraseq-strong(apebs, vtbl, q0) → valid-eraseq(q0) ⊦ valid-eraseq(eqent + q0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
eqent.pnum < # apebs, valid-eraseq-strong(apebs, vtbl, q0), (∃ apebs, vtbl. valid-eraseq-strong(apebs, vtbl, q0)) → valid-eraseq(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) ⊦ valid-eraseq(eqent ' + q0), eqent.pnum ∈ q0

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);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%a: ⊦ (q + q0) + q1 = q + q0 + q1; from specification list[erasequeue-base]% ⊦ eqent + q = eqent ' + q; (s)