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Proof step 1 from proof labsp (conclusion)

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

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ 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)
Proof step 7:
∀ 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)

Rule argument

indvar: i

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%int generated by 0, +1, -1;;%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs