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Proof step 9 from proof i2n-mod (previous step)

0 ≤ i0 ∧ 0 ≤ j, j0 < j, 0 < j, 0 ≤ j0, 0 ≤ i0 ⊦ i→n(j0) = (i→n(i0) * i→n(j) + i→n(j0)) % i→n(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mod-mult: n ≠ 0 ⊦ (m + n0 * n) % n = m % n; from specification nat-div%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs