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Proof step 11 from proof add (previous step)

0 ≤ j, 0 ≤ i1, 0 ≤ j → (i→n(i1) + 1) + i→n(j) = i→n((i1 + 1) + j), ∀ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
(i→n(i1) + 1) + i→n(j) = i→n((i1 + 1) + j), 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: zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs