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Proof step 6 from proof gcd-mod-one-01 (previous step)

(i * i0) % j = 1, gcd(i, j) = 1, 0 < j, 1 < j, 0 < i, 1 < i, i0 ≤ 0, i0 < j, i0 < i ⊦

User applied rule insert spec-lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: mod-ngr seq: ⊦ i < 0 ∧ 0 < j → (0 ≤ i % j ↔ i % j = 0) subst: [i, j] → [i * i0, j] flags:

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 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%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%zero-left: ⊦ 0 * i = 0; from specification int-mult%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%mod-zero-04: j ≠ 0 ⊦ 0 % j = 0; from specification int-div Used terms: i * i0, j