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Proof step 29 from proof keys-shift-left (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: n0 + 1

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n0 + 1