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Proof step 3 from proof le-sum (previous step)

n ≤ n0, n1 ≤ m0 + 1, n0 + m0 + 1 < n + n1, ∀ n, n0, n1. n ≤ n0 ∧ n1 ≤ m0 → n + n1 ≤ n0 + m0 ⊦ n + n1 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ (m + n0) + n1 = m + n0 + n1
Proof step 5:
n ≤ n0, n1 ≤ m0 + 1, m0 + 1 + n0 < n + n1, ∀ n, n0, n1. n ≤ n0 ∧ n1 ≤ m0 → n + n1 ≤ n0 + m0 ⊦ n + n1 = 0

Rule argument

spec: nat-basic2 inst: name: a seq: ⊦ (m + n0) + n1 = m + n0 + n1 subst: [n1, n0, m] → [n0, 1, m0] flags: rewrite

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: n0, 1, m0