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Proof step 7 from proof mod-mult-add (previous step)

0 < j, 0 < j, 0 < i0, ∀ i1, j, i. i→n(i1) < i→n(i0) → 0 < i1 ∧ 0 < j → (i + i1 * j) |%| j = i |%| j ⊦ (i + i0 * j - j) |%| j = i |%| j

User applied rule apply induction

And obtained the following new goals

Proof step 8:
(i + (i0 - 1) * j) |%| j = i |%| j, 0 < j, 0 < j, 0 < i0, ∀ i1, j, i. i→n(i1) < i→n(i0) → 0 < i1 ∧ 0 < j → (i + i1 * j) |%| j = i |%| j ⊦ (i + i0 * j - j) |%| j = i |%| j

Rule argument

subst: [i1, j, i] → [i0 - 1, j, i]

Additional information

Used simplifier rules: sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%one-01: ⊦ i = 1 → i * j = j; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%sub-add: ⊦ i + j - i = j; from specification int-abs%abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: i0 - 1, j, i