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Proof step 1 from proof mod-zero-02 (conclusion)

n ≠ 0 ⊦ 0 ≤ i → ¬ i % n→i(n) < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n→i(n) ≤ i, 0 ≤ i, i % n→i(n) < 0, 0 < i, i % n→i(n) + 1 < i ⊦ n = 1, n = 0

Rule argument

Additional information

Used simplifier rules: nat2int-02: ⊦ n→i(n) = 0 ↔ n = 0;%abs-02: ⊦ abs(n→i(n)) = n→i(n);%num-n2i-eq: 0 ≤ i ⊦ n→i(n) = (* i) ↔ n = i→n(i);%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) 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%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