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Proof step 13 from proof dom-max-in (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 14:
(dom st0).max < n, m < (dom st0).max, n ∈ st0[m, d] ⊦ st0 = ∅, ∃ n. (dom st0).max < n ∧ n ∈ st0
Proof step 16:
m < n, n ∈ st0[m, d] ⊦ st0 = ∅, m < (dom st0).max, ∃ n. (dom st0).max < n ∧ n ∈ st0

Rule argument

pos: left 1

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