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Proof step 2 from proof labsp (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 3:
0 ≤ i0 +1, 0 ≤ j, ∀ j. 0 ≤ j ∧ 0 ≤ i0 → (i→n(i0) < i→n(j) ↔ abs(i0) < abs(j)) ⊦ i→n(i0 +1) < i→n(j) ↔ abs(i0 +1) < abs(j)
Proof step 8:
0 ≤ i0 -1, 0 ≤ j, ∀ j. 0 ≤ j ∧ 0 ≤ i0 → (i→n(i0) < i→n(j) ↔ abs(i0) < abs(j)) ⊦ i→n(i0 -1) < i→n(j) ↔ abs(i0 -1) < abs(j)

Rule argument

indvar: i

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%int generated by 0, +1, -1;;% ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ 0 ≤ i → abs(i) = i; (s)% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)% ⊦ i ≤ j → ¬ j < i; (s)