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Proof step 13 from proof valid-eraseq-strong-imp-p (previous step)

apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m0 < # apebs, valid-eraseq(apebs, q0), valid-eraseq-strong(apebs, vtbl, q0), ∀ v, m. (valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ↔ lref = v × m) ∧ (lref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → avhdr.sqn ≠ valid-sqns(v, m, apebs).max) ⊦ m0 ∈ q0, eq-entry(m0, lref).pnum < # apebs ∧ ( valid-peb(apebs[eq-entry(m0, lref).pnum]) → eq-entry(m0, lref).lebref = apebs[eq-entry(m0, lref).pnum].vidhdr.vol × apebs[eq-entry(m0, lref).pnum].vidhdr.leb) ∧ valid-eraseq(apebs, q0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m0 < # apebs, valid-eraseq(apebs, q0), valid-eraseq-strong(apebs, vtbl, q0), valid-peb(mkpeb(aehdr, avhdr, buf, m1, boolvar)), ∀ v, m. (valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ↔ lref = v × m) ∧ (lref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → avhdr.sqn ≠ valid-sqns(v, m, apebs).max) ⊦ lref = avhdr.vol × avhdr.leb, m0 ∈ q0

Rule argument

Additional information

Used simplifier rules: ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)% ⊦ eq-entry(m, lref).lebref = lref; (s)% ⊦ eq-entry(m, lref).pnum = m; (s)