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Proof step 6 from proof lf-valid-recs-sqns-vidhdr (previous step)

apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), recs[v × m] = recovery-entry(n0, n1), v × m ∈ recs, ∀ ladr. ladr ∈ recs → recs[ladr].pnum < # apebs ∧ apebs[recs[ladr].pnum].vidhdr.sqn = recs[ladr].sqn ⊦ avhdr.sqn = n1

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: v × m

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ recovery-entry(m, m0).pnum = m; (s)% ⊦ recovery-entry(m, m0).sqn = m0; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s) Used terms: v × m