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Proof step 1 from proof abs-neg (conclusion)

⊦ i→n(~ i) = i→n(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
0 ≤ i ⊦ i→n(~ i) = i→n(i)
Proof step 3:
¬ 0 ≤ i ⊦ i→n(~ i) = i→n(i)

Rule argument

spec: intnat inst: name: abs-negp seq: 0 ≤ i ⊦ i→n(~ i) = i→n(i) subst: [i] → [i] flags: rewrite, current

Additional information

Used terms: i