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Proof step 3 from proof inni-ls-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n→i(m + n) = n→i(m) + n→i(n)
Proof step 5:
0 < j, n→i(m + n) < j, i→n(j) ≤ m + n ⊦

Rule argument

spec: <toplevel> name: n2i-add seq: ⊦ n→i(m + n) = n→i(m) + n→i(n) subst: [n, m] → [n, m] flags: <rotated> <all paths>

Additional information

Used terms: n, m