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Proof step 4 from proof add (previous step)

~1 ≤ i0, 0 ≤ j, ∀ j. 0 ≤ j ∧ 0 ≤ i0 → i→n(i0) + i→n(j) = i→n(i0 + j) ⊦ i→n(i0 + 1) + i→n(j) = i→n((i0 + 1) + j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < i → i→n(i) = i→n(i - 1) + 1
Proof step 6:
0 < i0 + 1, 0 ≤ j, ~1 ≤ i0, ∀ j. 0 ≤ j ∧ 0 ≤ i0 → i→n(i0) + i→n(j) = i→n(i0 + j) ⊦ (i→n(i0 + 1 - 1) + 1) + i→n(j) = i→n((i0 + 1) + j)

Rule argument

spec: <toplevel> name: Int2nat-pos seq: ⊦ 0 < i → i→n(i) = i→n(i - 1) + 1 subst: [i] → [i0 + 1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ 0 + m = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ i + 0 = i; (s) Used terms: i0 + 1