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Proof step 1 from proof mod-lsz (conclusion)

⊦ 0 ≤ j ∧ 0 < i → ¬ j % 2 ^ i < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
2 ^ i ≤ j, j % 2 ^ i < i, j % 2 ^ i + 1 < i, 0 ≤ j, 0 < i, j % 2 ^ i < 0, 0 < j, j % 2 ^ i + 1 < j ⊦ i = 0

Rule argument

Additional information

Used simplifier rules: zero: ⊦ 0 < j → (i ^ j = 0 ↔ i = 0);%abs-2pot: ⊦ abs(2 ^ i) = 2 ^ i;%eq-one: ⊦ 2 ^ i = 1 ↔ i = 0;%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%mod-ls-same-01: j ≠ 0 ⊦ i % j < i ↔ 0 < i ∧ (abs(j) = 1 ∨ abs(j) ≤ i); from specification int-div%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%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%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs