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Proof step 18 from proof in-keys (previous step)

m + nused0 < # zbrar, m < # zbrar, n < nused0, zbrar[(m + 1) + n].key ∈ keys(zbrar, m + 1, nused0) ⊦ # keys(zbrar, m + 1, nused0) = 0, keys(zbrar, m + 1, nused0) = ∅, ∃ n0. ¬ nused0 < n0 ∧ zbrar[m + n0].key = zbrar[(m + 1) + n].key

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: 1 + n

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: 1 + n