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Proof step 2 from proof le-add-01 (previous step)

n < m, n0 < m0, m + m0 ≤ n + n0 ⊦ m = 0, m0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0)
Proof step 4:
n < m, ∃ m1. n0 + m1 = m0 ∧ m1 ≠ 0, m + m0 ≤ n + n0 ⊦ m = 0, m0 = 0

Rule argument

spec: nat-basic2 inst: name: addex seq: ⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0) subst: [n, m] → [m0, n0] flags: rewrite

Additional information

Used terms: m0, n0