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Proof step 1 from proof rax-02 (conclusion)

⊦ i + 1 < 0 → i→n(i + 1) = i→n(i) -1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ i < 0 → i→n(i) = i→n(~ i)
Proof step 3:
i < 0 ⊦ i + 1 < 0 → i→n(i + 1) = i→n(~ i) -1

Rule argument

spec: intnat inst: name: Int2nat-neg seq: ⊦ i < 0 → i→n(i) = i→n(~ i) subst: [i] → [i] flags: rewrite, current

Additional information

Used simplifier rules: rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;%pred-03: ⊦ n -1 = n - 1; from specification nat%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%zero-gels: ⊦ i < 0 ∧ 0 ≤ j → ¬ j < i; from specification int-abs Used terms: i