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Proof step 6 from proof max-plus-one (previous step)

ns.max + 1 ∈ ns ⊦ # ns = 0, ns = ∅, ns.max < ns.max + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0)

Rule argument

spec: nat-basic name: addex seq: ⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0) subst: [n, m] → [ns.max + 1, ns.max] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: ns.max + 1, ns.max