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Proof step 3 from proof i2i-neg (previous step)

i < 0 ⊦ i + n→i(i→n(i)) = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < i → n→i(i→n(i)) = i

Rule argument

spec: <toplevel> name: i2i-pos seq: ⊦ 0 < i → n→i(i→n(i)) = i subst: [i] → [~ i]

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i);% ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ ~ i = j ↔ i + j = 0; (s) Used terms: ~ i