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Proof step 2 from proof pos-add (previous step)

n→i(m) + n→i(n) < j, n→i(m) + n→i(n) < 0, j ≤ 0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ n→i(m + n) = n→i(m) + n→i(n)
Proof step 4:
n→i(m + n) < j, n→i(m + n) < 0, j ≤ 0 ⊦

Rule argument

spec: intnat inst: name: n2i-add seq: ⊦ n→i(m + n) = n→i(m) + n→i(n) subst: [n, m] → [n, m] flags: rewrite, current

Additional information

Used terms: n, m