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Proof step 9 from proof in-payload (previous step)

key ∈ payload(zbrar, m + 1, nused0) ↔ (∃ n. n < nused0 ∧ zbrar[(m + 1) + n].entry? ∧ zbrar[(m + 1) + n].key = key), zbrar[m].entry?, m + nused0 < # zbrar, m < # zbrar ⊦ key ∈ payload(zbrar, m + 1, nused0)[zbrar[m].key, zbrar[m].adr] ↔ (∃ n. ¬ nused0 < n ∧ zbrar[m + n].entry? ∧ zbrar[m + n].key = key)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
key ∈ payload(zbrar, m + 1, nused0) ↔ (∃ n. n < nused0 ∧ zbrar[(m + 1) + n].entry? ∧ zbrar[(m + 1) + n].key = key), zbrar[m] = mkzentry(key0, adr), zbrar[m].entry?, m < # zbrar, m + nused0 < # zbrar ⊦ key ∈ payload(zbrar, m + 1, nused0)[key0, adr] ↔ (∃ n. ¬ nused0 < n ∧ zbrar[m + n].entry? ∧ zbrar[m + n].key = key)

Rule argument

spec: zbranch name: elim-03 seq: ⊦ zbr.entry? → (key = zbr.key ∧ adr = zbr.adr ↔ zbr = mkzentry(key, adr)) subst: [zbr, key, adr] → [zbrar[m], key0, adr]

Additional information

Used simplifier rules: ⊦ zbr.entry? → (key = zbr.key ∧ adr = zbr.adr ↔ zbr = mkzentry(key, adr)); (elim) Used terms: zbrar[m], key0, adr