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Proof step 6 from proof add (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 ≤ i0, 0 ≤ j, ∀ j0. 0 ≤ j0 → i→n(i0) + i→n(j0) = i→n(i0 + j0) ⊦ (i→n(i0) + 1) + i→n(j) = i→n((i0 + 1) + j)

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + 0 = i; (s)% ⊦ (* i) + i0 - (* j) = (i - j) + i0; (s)