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Proof step 20 from proof valid-eraseq-strong-notinvols (previous step)

apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), n < # apebs, m0 < # 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, 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), ¬ v ∈ vtbl ∨ ¬ m < vtbl[v] ⊦ n = m0, valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) '), n ∈ q0, m0 ∈ q0

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: n, apebs, vtbl, v, m

Additional information

Used simplifier rules: ⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s) Used terms: n, apebs, vtbl, v, m