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Proof step 10 from proof add (previous step)

i0 = i1 + 1, 0 < i0, 0 ≤ j, ∀ j. 0 ≤ j ∧ 0 ≤ i0 → i→n(i0) + i→n(j) = i→n(i0 + j) ⊦ i→n(i1) + i→n(j) = i→n(j + i0 - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ j, 0 ≤ i1, ∀ j0. 0 ≤ j0 → (i→n(i1) + 1) + i→n(j0) = i→n((i1 + 1) + j0) ⊦ i→n(i1) + i→n(j) = i→n(j + i1)

Rule argument

Additional information

Used simplifier rules: rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; (s)% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (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)