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Proof step 5 from proof rounded-help2 (previous step)

n0 < m2, m1 < m2, 2 * m2 ≤ n0 + m1 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0)
Proof step 7:
n0 < m2, 2 * m2 ≤ n0 + m1, ∃ m0. m1 + m0 = m2 ∧ m0 ≠ 0 ⊦

Rule argument

spec: nat-basic name: addex seq: ⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0) subst: [n, m] → [m2, m1] flags: <all paths>

Additional information

Used terms: m2, m1