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Proof step 15 from proof in-keys (previous step)

n = n1 +1, m + nused0 < # zbrar, m < # zbrar, n ≤ nused0 ⊦ zbrar[m + n].key = zbrar[m].key, zbrar[m + n].key ∈ keys(zbrar, m + 1, nused0), ∃ n0. n0 < nused0 ∧ zbrar[(m + 1) + n0].key = zbrar[m + n].key

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
n1 < nused0, m + nused0 < # zbrar, m < # zbrar ⊦ zbrar[m + n1 + 1].key = zbrar[m].key, zbrar[m + n1 + 1].key ∈ keys(zbrar, m + 1, nused0), ∃ n. n < nused0 ∧ zbrar[(m + 1) + n].key = zbrar[m + n1 + 1].key

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)