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Proof step 103 from proof payload-shift-left (previous step)

(n0 < n ⊃ zbrar[n0];zbrar[n0 + 1]) = mkzentry(key, adr0), zbrar[n2 + 1] = mkzentry(key, adr), (n0 < n ⊃ zbrar[n0];zbrar[n0 + 1]).entry?, 2 + n0 < # zbrar, n + 1 < # zbrar, n0 + 1 < nused, n0 < # zbrar, n0 < nused, n < nused, n0 < n2, n2 < # zbrar, n2 + 1 < # zbrar, n2 < nused, nused < # zbrar, n < # zbrar, n0 + 1 < # zbrar, sorted(zbrar, 0, nused + 1), key ∈ payload(shl(zbrar, n, nused + 1), 0, nused), n ≤ n2 + 1, n ≤ n2, ∀ n1. ¬ n2 < n1 ∧ zbrar[n1].entry? → zbrar[n1].key ≠ key ⊦ payload(shl(zbrar, n, nused + 1), 0, nused)[key] = adr, zbrar[n].entry?

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 104:
(n0 < n ⊃ zbrar[n0];zbrar[n0 + 1]) = mkzentry(key, adr0), zbrar[n2 + 1] = mkzentry(key, adr), 2 + n0 < # zbrar, n + 1 < # zbrar, n0 + 1 < nused, n0 < # zbrar, n0 < nused, n < nused, n0 < n2, n2 < # zbrar, n2 + 1 < # zbrar, n2 < nused, nused < # zbrar, n < # zbrar, n0 + 1 < # zbrar, sorted(zbrar, 0, nused + 1), key ∈ payload(shl(zbrar, n, nused + 1), 0, nused), n ≤ n2 + 1, n ≤ n2, ∀ n1. ¬ n2 < n1 ∧ zbrar[n1].entry? → zbrar[n1].key ≠ key ⊦ payload(shl(zbrar, n, nused + 1), 0, nused)[key] = adr, zbrar[n].entry?

Rule argument

Additional information

Used simplifier rules: ⊦ mkzentry(key, adr).entry?; (s)