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Proof step 1 from proof nat2int-09 (conclusion)

⊦ i→n(n→i(n) + 1) = n + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1

Rule argument

spec: intnat inst: name: rax-04 seq: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1 subst: [i] → [n→i(n)] flags: rewrite, current

Additional information

Used simplifier rules: add-nonneg-int-to-nat: ⊦ 0 ≤ i → i→n(i + n→i(n)) = i→n(i) + n;%n2n: ⊦ i→n(n→i(n)) = n;%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: n→i(n)