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Proof step 16 from proof dom-max-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
m < n, (dom st0).max < n, n ∈ st0[m, d], (dom st0).max ≤ m ⊦ st0 = ∅, ∃ n. (dom st0).max < n ∧ n ∈ st0

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)