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Proof step 11 from proof labs (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 12:
0 ≤ i, 0 ≤ j ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j)
Proof step 13:
⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j), i ≤ 0 ∧ 0 ≤ j, 0 ≤ i ∧ j ≤ 0, 0 ≤ i ∧ 0 ≤ j, 0 ≤ ~ i ∧ 0 ≤ ~ j
Proof step 15:
i→n(~ i) < i→n(~ j) ↔ abs(~ i) < abs(~ j) ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j), i ≤ 0 ∧ 0 ≤ j, 0 ≤ i ∧ j ≤ 0, 0 ≤ i ∧ 0 ≤ j

Rule argument

spec: <toplevel> name: labsp seq: 0 ≤ i, 0 ≤ j ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j) subst: [i, j] → [~ i, ~ j]

Additional information

Used terms: ~ i, ~ j