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Proof step 2 from proof diff-primes-dups (previous step)

prime(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))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
∀ i. i ∈ ints → 0 ≤ i ∧ i < j, 0 ≤ j, j < n→i(# ints) ⊦ dups(ints)
Proof step 4:
¬ ( (∀ i. i ∈ countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) + countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j) → 0 ≤ i ∧ i < j) ∧ 0 ≤ j ∧ j < n→i(#(countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) + countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j)))), prime(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))

Rule argument

spec: intlist inst: name: range-dups seq: ∀ i. i ∈ ints → 0 ≤ i ∧ i < j, 0 ≤ j, j < n→i(# ints) ⊦ dups(ints) subst: [j, ints] → [j, countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) + countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j)] flags: rewrite

Additional information

Used terms: j, countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) + countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j)