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Proof step 3 from proof valid-recs-sqns-update (previous step)

∀ ladr. ladr ∈ recs → recs[ladr].pnum < # apebs ∧ apebs[recs[ladr].pnum].vidhdr.sqn = recs[ladr].sqn ⊦ (∀ ladr. ladr ∈ recs[v × m, recovery-entry(m0, m1)] → recs[v × m, recovery-entry(m0, m1)][ladr].pnum < # apebs ∧ apebs[recs[v × m, recovery-entry(m0, m1)][ladr].pnum].vidhdr.sqn = recs[v × m, recovery-entry(m0, m1)][ladr].sqn) ↔ m0 < # apebs ∧ apebs[m0].vidhdr.sqn = m1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
apebs[m0] = mkpeb(aehdr, avhdr, buf, m2, boolvar), ∀ ladr. ladr ∈ recs → recs[ladr].pnum < # apebs ∧ apebs[recs[ladr].pnum].vidhdr.sqn = recs[ladr].sqn ⊦ (∀ ladr. ladr ∈ recs[v × m, recovery-entry(m0, m1)] → recs[v × m, recovery-entry(m0, m1)][ladr].pnum < # apebs ∧ apebs[recs[v × m, recovery-entry(m0, m1)][ladr].pnum].vidhdr.sqn = recs[v × m, recovery-entry(m0, m1)][ladr].sqn) ↔ m0 < # apebs ∧ avhdr.sqn = m1

Rule argument

spec: apeb name: elim seq: ⊦ ( ( (aehdr = apeb.echdr ∧ avhdr = apeb.vidhdr) ∧ buf = apeb.data) ∧ m = apeb.fill) ∧ (boolvar ↔ apeb.bad) ↔ apeb = mkpeb(aehdr, avhdr, buf, m, boolvar) subst: [apeb, aehdr, avhdr, buf, m, boolvar] → [apebs[m0], aehdr, avhdr, buf, m2, boolvar]

Additional information

Used simplifier rules: ⊦ ( ((aehdr = apeb.echdr ∧ avhdr = apeb.vidhdr) ∧ buf = apeb.data) ∧ m = apeb.fill) ∧ (boolvar ↔ apeb.bad) ↔ apeb = mkpeb(aehdr, avhdr, buf, m, boolvar); (elim) Used terms: apebs[m0], aehdr, avhdr, buf, m2, boolvar