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Proof step 7 from proof diff-primes-dups (previous step)

0 ≤ 0 ∧ 0 ≤ (j - 1) / 2, 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(i→n(1 + (j - 1) / 2 - 0)) + 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
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 < abs(1 + (j - 1) / 2) + 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: ini2-abs: ⊦ n→i(i→n(i)) = abs(i); from specification intnat%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%div-two-zero-01: ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i; from specification int-div