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Proof step 4 from proof diff-primes-dups (previous step)

¬ ( (∀ 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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
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 < n→i(# countFun(0, (j - 1) / 2, λ i. i ^2 |%| j)) + n→i(# countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j)) ∧ (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

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%prime-not-neg: ⊦ j < 0 → ¬ prime(j); from specification int-gcd%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%n2i-add: ⊦ n→i(m + n) = n→i(m) + n→i(n); from specification intnat%len: ⊦ #(ints + ints0) = # ints + # ints0; from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%in-append: ⊦ i ∈ ints + ints0 ↔ ¬ ¬(i ∈ ints ∨ i ∈ ints0); from specification list[intlista]