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Proof step 4 from proof l-03 (previous step)

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

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 5:
⊦ n < n0 ∧ n0 < n1 → n < n1

Rule argument

spec: nat-basic1 inst: name: trans-1 seq: ⊦ n < n0 ∧ n0 < n1 → n < n1 subst: [n0, n, n1] → [m0 + n, m + n, m0 + n0] flags:

Additional information

Used simplifier rules: l: ⊦ m + n < m + n0 ↔ n < n0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m; Used terms: m0 + n, m + n, m0 + n0