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Proof step 5 from proof le-sum (previous step)

n ≤ n0, n1 ≤ m0 + 1, m0 + 1 + n0 < 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 6:
⊦ m + n = n + m
Proof step 7:
n ≤ n0, n1 ≤ m0 + 1, m0 + n0 + 1 < n + n1, ∀ n, n0, n1. n ≤ n0 ∧ n1 ≤ m0 → n + n1 ≤ n0 + m0 ⊦ n + n1 = 0

Rule argument

spec: nat-basic2 inst: name: c seq: ⊦ m + n = n + m subst: [n, m] → [n0, 1] flags: rewrite

Additional information

Used terms: n0, 1