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Proof step 59 from proof payload-shift-left (previous step)

zbrar[n] = mkzentry(key, adr), zbrar[n0].entry?, n < # zbrar, nused < # zbrar, n0 < # zbrar, sorted(zbrar, 0, nused + 1), n ≤ nused, n0 ≤ nused ⊦ zbrar[n0].key = key, ∃ n1. n1 < nused ∧ (n1 < n ⊃ zbrar[n1];zbrar[n1 + 1]).entry? ∧ (n1 < n ⊃ zbrar[n1];zbrar[n1 + 1]).key = zbrar[n0].key

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 60:
zbrar[n] = mkzentry(key, adr), zbrar[n0] = mkzentry(key0, adr0), zbrar[n0].entry?, n0 < # zbrar, nused < # zbrar, n < # zbrar, sorted(zbrar, 0, nused + 1), n0 ≤ nused, n ≤ nused ⊦ key0 = key, ∃ n1. n1 < nused ∧ (n1 < n ⊃ zbrar[n1];zbrar[n1 + 1]).entry? ∧ (n1 < n ⊃ zbrar[n1];zbrar[n1 + 1]).key = key0

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[n0], key0, adr0]

Additional information

Used simplifier rules: ⊦ zbr.entry? → (key = zbr.key ∧ adr = zbr.adr ↔ zbr = mkzentry(key, adr)); (elim) Used terms: zbrar[n0], key0, adr0