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Proof step 36 from proof keys-shift-left (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 37:
n0 = m + 1, n < # zbrar, nused < # zbrar, n0 < # zbrar, sorted(zbrar, 0, nused + 1), n0 ≤ nused, n ≤ n0, n ≤ nused ⊦ zbrar[n0].key = zbrar[n].key, n0 = 0, ∃ n1. n1 < nused ∧ (n1 < n ⊃ zbrar[n1];zbrar[n1 + 1]).key = zbrar[n0].key, m < nused ∧ (m < n ⊃ zbrar[m];zbrar[m + 1]).key = zbrar[n0].key

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n0, m]

Additional information

Used simplifier rules: ⊦ m ≤ 0 ↔ m = 0; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n0, m