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Proof step 7 from proof labs (previous step)

⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j), 0 ≤ i ∧ 0 ≤ j, 0 ≤ i ∧ 0 ≤ ~ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j), 0 ≤ i ∧ 0 ≤ j, 0 ≤ i ∧ j ≤ 0

Rule argument

Additional information

Used simplifier rules: 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 ≤ ~ j ↔ i + j ≤ 0; (s)