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Proof step 2 from proof labsp (previous step)

∀ j. 0 ≤ i0 ∧ 0 ≤ j → (i→n(i0) < i→n(j) ↔ abs(i0) < abs(j)), 0 ≤ i0 +1, 0 ≤ j ⊦ i→n(i0 +1) < i→n(j) ↔ abs(i0 +1) < abs(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
~1 ≤ i0, 0 ≤ j, ∀ j0. 0 ≤ i0 ∧ 0 ≤ j0 → (i→n(i0) < i→n(j0) ↔ i0 < j0) ⊦ i→n(i0 + 1) < i→n(j) ↔ abs(i0 + 1) < j

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2