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Proof step 2 from proof pot-lez (previous step)

0 < i ⊦ i ^ j ≤ 0 ↔ i ^ j = 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: notneg-02: ⊦ 0 < i → ¬ i ^ j < 0;%pot-zero-01: ⊦ 0 < i → (i ^ j = 0 ↔ j < 0 ∧ 1 < i);%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;% ⊦ 1 < i → abs(i) = i; (s)%forward-le-02: ⊦ 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%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%j ≠ 0 ⊦ 0 < 1 / j ↔ j = 1; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)