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Proof step 5 from proof labs (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 6:
0 ≤ i, 0 ≤ j ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j)
Proof step 7:
⊦ 0 ≤ i ∧ 0 ≤ ~ j, i→n(i) < i→n(j) ↔ abs(i) < abs(j), 0 ≤ i ∧ 0 ≤ j

Rule argument

spec: intnat inst: name: labsp seq: 0 ≤ i, 0 ≤ j ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j) subst: [i, j] → [i, ~ j] flags: current, []

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i);%abs-neg: ⊦ abs(~ i) = abs(i); from specification int-abs Used terms: i, ~ j