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

i→n(~ i) = 0, i < 0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < i → i→n(i) = i→n(i - 1) + 1

Rule argument

spec: intnat inst: name: Int2nat-pos seq: ⊦ 0 < i → i→n(i) = i→n(i - 1) + 1 subst: [i] → [~ i] flags: rewrite, current

Additional information

Used simplifier rules: oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: ~ i