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Proof step 15 from proof dom-max-in (previous step)

m < n, (dom st0).max < n, m < (dom st0).max, m + 1 < n, n ∈ st0[m, d] ⊦ st0 = ∅, ∃ n. (dom st0).max < n ∧ n ∈ st0

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: n

Additional information

Used simplifier rules: ⊦ n ∈ st[n0, d] ↔ n = n0 ∨ n ∈ st; (ws)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + m0 < n → ¬ n + n0 < m; (s) Used terms: n