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Proof step 3 from proof rax-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

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: abs-neg: ⊦ i→n(~ i) = i→n(i);%abs-neg: ⊦ i→n(~ i) = i→n(i);%subsub: ⊦ ~ i - j = ~(i + j); from specification int-abs%pred: ⊦ (m + 1)-1 = m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%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%neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs Used terms: ~ i