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Proof step 18 from proof mod-mult-add (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 19:
i0 < j, i0 ≤ 0, 0 < j, ∀ i1, j, i. i→n(i1) < i→n(i0) → 0 < j ∧ i1 ≤ 0 ∧ i1 < j → (i + i1 * j) |%| j = i |%| j ⊦ 0 < j ∧ i0 + 1 ≤ 0 ∧ i0 + 1 < j, (i + i0 * j) |%| j = i |%| j
Proof step 21:
(i + (i0 + 1) * j) |%| j = i |%| j, i0 < j, i0 ≤ 0, 0 < j, ∀ i1, j, i. i→n(i1) < i→n(i0) → 0 < j ∧ i1 ≤ 0 ∧ i1 < j → (i + i1 * j) |%| j = i |%| j ⊦ (i + i0 * j) |%| j = i |%| j

Rule argument

subst: [i1, j, i] → [i0 + 1, j, i]

Additional information

Used simplifier rules: abs-succ-01: ⊦ i→n(i + 1) < i→n(i) ↔ i < 0; from specification intnat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = 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%Mult-zero: ⊦ i * 0 = 0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%ax-04: ⊦ i + 0 = i; from specification int-basic2 Used terms: i0 + 1, j, i