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Proof step 5 from proof abs-pred-01 (previous step)

i→n(i) = m + 1, i < 0 ⊦ i = ~1, i = 0, i→n(i + 1) = m

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i < 0 → i→n(i) = i→n(~ i)
Proof step 7:
i→n(~ i) = m + 1, i < 0 ⊦ i = 0, i = ~1, i→n(i + 1) = m

Rule argument

spec: <toplevel> name: Int2nat-neg seq: ⊦ i < 0 → i→n(i) = i→n(~ i) subst: [i] → [i] flags: <all paths>

Additional information

Used terms: i