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Proof step 2 from proof Shor-algorithm (previous step)

gcd(i, j) = 1, gcd(i ^ (i0 / 2) - 1, j) < i, gcd(i ^ (i0 / 2) - 1, j) < j, 0 ≤ i0, 0 < i, 1 < i, 0 < j, 1 < j, gcd(i ^ (i0 / 2) - 1, j) ≤ 1, even(i0), ≡mod(i ^ i0, 1, j), i < j ⊦ prime(j), ≡mod(i ^ (i0 / 2), ~1, j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i ≠ j → (i ≤ j ↔ i ≤ j - 1)
Proof step 4:
¬ gcd(i ^ (i0 / 2) - 1, j) ≠ 1, gcd(i, j) = 1, gcd(i ^ (i0 / 2) - 1, j) < i, gcd(i ^ (i0 / 2) - 1, j) < j, 0 ≤ i0, 0 < i, 1 < i, 0 < j, 1 < j, gcd(i ^ (i0 / 2) - 1, j) ≤ 1, even(i0), ≡mod(i ^ i0, 1, j), i < j ⊦ prime(j), ≡mod(i ^ (i0 / 2), ~1, j)

Rule argument

spec: int-abs inst: name: le-rec seq: ⊦ i ≠ j → (i ≤ j ↔ i ≤ j - 1) subst: [j, i] → [1, gcd(i ^ (i0 / 2) - 1, j)] flags: rewrite

Additional information

Used simplifier rules: gcd-less-eq-zero: ⊦ gcd(i, j) ≤ 0 ↔ gcd(i, j) = 0; from specification int-gcd%gcd-zero-zero: ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0; from specification int-gcd%sub-eq: ⊦ 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%sub-eq-01: ⊦ i = j → i - j = 0; from specification int-abs%gcd-zero-04: ⊦ gcd(i, 0) = abs(i); from specification int-gcd%abs: ⊦ 1 < i → abs(i) = i; from specification int-abs%gcd-com: ⊦ gcd(i, j) = gcd(j, i); from specification int-gcd%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: 1, gcd(i ^ (i0 / 2) - 1, j)