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Proof step 18 from proof abs-ls-same-01 (previous step)

⊦ 0 < i + abs(i) ↔ ¬ i ≤ 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i + j < j ↔ i < 0; (s)% ⊦ i < j → ¬ j < i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < i + j ↔ 0 < j; (s)% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i + i = 0; (s)