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Proof step 21 from proof mod-pot-shift-pos (previous step)

0 < j, ∀ i1, j, i. i→n(i1) < i→n(0 + 1) → 0 < j ∧ 0 < i1 → (i % j) ^ i1 % j = i ^ i1 % j ∨ i1 = 0 ⊦ (1 % j * i % j) % j = i % j, 0 = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ 1 < j → 1 % j = 1

Rule argument

spec: int-div name: one-03 seq: ⊦ 1 < j → 1 % j = 1 subst: [j] → [j] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 1 * i = i; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%j ≠ 0 ⊦ (i % j) % j = i % j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s)% ⊦ i = 1 → j % i = 0; (s)%j ≠ 0 ⊦ 0 % j = 0; (s) Used terms: j