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Proof step 16 from proof diff-primes-dups (previous step)

prime(j), 0 ≤ j ⊦ j = 2, dups(countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) + countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j)), dups(countFun(0, (j - 1) / 2, λ i. i ^2 |%| j)), dups(countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j)), j < 2 * 1 + 2 * (j - 1) / 2 ∧ (i ∈ countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) ∨ i ∈ countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j) → 0 ≤ i ∧ i < j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
j ≠ 0 ⊦ i % j = 0 → i / j * j = i
Proof step 18:
2 ≠ 0 ∧ (j - 1) % 2 = 0, prime(j), 0 ≤ j ⊦ j = 2, dups(countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) + countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j)), dups(countFun(0, (j - 1) / 2, λ i. i ^2 |%| j)), dups(countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j)), j < 2 * 1 + (j - 1) ∧ (i ∈ countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) ∨ i ∈ countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j) → 0 ≤ i ∧ i < j)

Rule argument

spec: int-div inst: name: divmult seq: j ≠ 0 ⊦ i % j = 0 → i / j * j = i subst: [j, i] → [2, j - 1] flags: rewrite

Additional information

Used simplifier rules: associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%prime-odd-01: ⊦ prime(j) ∧ j ≠ 2 → (j - 1) % 2 = 0; from specification int-gcd%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 2, j - 1