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Proof step 3 from proof le-add-02 (previous step)

n < m, n0 < m0, m + m0 ≤ n + n0 + 1 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ m ≤ n ↔ (∃ n0. m + n0 = n)
Proof step 5:
n0 < m0, n < m, ∃ n1. (m + m0) + n1 = n + n0 + 1 ⊦

Rule argument

spec: <toplevel> name: le-addex seq: ⊦ m ≤ n ↔ (∃ n0. m + n0 = n) subst: [n, m] → [n + n0 + 1, m + m0] flags: <all paths>

Additional information

Used terms: n + n0 + 1, m + m0