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Proof step 8 from proof dom-max-in (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 9:
st0 = ∅, m < n, n ∈ st0[m, d], (∃ n. (dom st0).max < n ∧ n ∈ st0) → st0 = ∅ ⊦
Proof step 12:
(m < (dom st0).max ⊃ (dom st0).max;m) < n, n ∈ st0[m, d], (∃ n. (dom st0).max < n ∧ n ∈ st0) → st0 = ∅ ⊦ st0 = ∅

Rule argument

pos: left 1

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