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Proof step 14 from proof dom-max-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
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

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%transitivity: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification oelem[nat-set]%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic