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Proof step 14 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 + (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 15:
⊦ i * (j + j0) = i * j + i * j0
Proof step 16:
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)

Rule argument

spec: int-mult inst: name: dist-rsplit seq: ⊦ i * (j + j0) = i * j + i * j0 subst: [j0, j, i] → [(j - 1) / 2, 1, 2] flags: rewrite

Additional information

Used terms: (j - 1) / 2, 1, 2