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Proof step 14 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, i ≤ 0 ∧ j ≤ 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 5

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%zero-01: ⊦ ¬ m < 0; from specification nat-basic1%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs