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Proof step 2 from proof grz (previous step)

0 < i ⊦ 0 < i ^ j ↔ i = 1 ∨ 0 ≤ j

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: Pot-neg: ⊦ i < 0 → j ^ i = 1 / j ^ abs(i);%zero-02: ⊦ j < 0 → (i ^ ~ j = 0 ↔ i = 0);%eqone-02: ⊦ i ^ ~ j = 1 ∧ j < 0 → abs(i) = 1;%one-01: ⊦ 1 ^ j = 1;%notneg-01: ⊦ 0 < i ∧ 0 ≤ j → ¬ i ^ j ≤ 0;%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%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%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%one-pos: j ≠ 0 ⊦ 0 < 1 / j ↔ j = 1; from specification int-div%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs