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

i→n(i - 1) < i→n(i), i < 0 ⊦

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 < 0, i→n(~(i - 1)) < i→n(i), i - 1 < 0 ⊦

Rule argument

spec: <toplevel> name: Int2nat-neg seq: ⊦ i < 0 → i→n(i) = i→n(~ i) subst: [i] → [i - 1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i < j → ¬ j < i; (s) Used terms: i - 1