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Proof step 5 from proof dom-max-in (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ (ns ++ n).max = (ns = ∅ ⊃ n;(n < ns.max ⊃ ns.max;n))
Proof step 7:
(dom st0 = ∅ ⊃ m;(m < (dom st0).max ⊃ (dom st0).max;m)) < n, n ∈ st0[m, d], (∃ n. (dom st0).max < n ∧ n ∈ st0) → st0 = ∅ ⊦

Rule argument

spec: oset, inst: nat-set name: max-ifins seq: ⊦ (ns ++ n).max = (ns = ∅ ⊃ n;(n < ns.max ⊃ ns.max;n)) subst: [n, ns] → [m, dom st0] flags: <all paths>

Additional information

Used terms: m, dom st0