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Proof step 1 from proof mod2-pos (conclusion)

⊦ 0 ≤ i ∧ 0 ≤ j → ¬ j % 2 ^ i < 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: two-zero: ⊦ 0 ≤ j → 2 ^ j ≠ 0;%abs-2pot: ⊦ abs(2 ^ i) = 2 ^ i;%eq-one: ⊦ 2 ^ i = 1 ↔ i = 0;%mod-lsz: ⊦ 0 ≤ j ∧ 0 < i → ¬ j % 2 ^ i < 0;%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%mod-zero-09: ⊦ 0 < i ∧ i < j → i % j = i; from specification int-div%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%mod-one: ⊦ i % 1 = 0; from specification int-div%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%mod-ls-same-01: j ≠ 0 ⊦ i % j < i ↔ 0 < i ∧ (abs(j) = 1 ∨ abs(j) ≤ i); from specification int-div%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs