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Proof step 10 from proof diff-primes (previous step)

¬(n→i(m) ≤ (j - 1) / 2 - 0 ∧ 0 ≤ 0 ∧ 0 ≤ (j - 1) / 2), countFun(0, (j - 1) / 2, λ i. i ^2 |%| j)[m] = countFun(0, (j - 1) / 2, λ i. i ^2 |%| j)[n], 0 ≤ j, m < i→n(1 + (j - 1) / 2), n < i→n(1 + (j - 1) / 2), prime(j) ⊦ countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) = [], j = 2, m = n, ~1 = (j - 1) / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
countFun(0, (j - 1) / 2, λ i. i ^2 |%| j)[m] = countFun(0, (j - 1) / 2, λ i. i ^2 |%| j)[n], 0 ≤ j, m < i→n(1 + (j - 1) / 2), n < i→n(1 + (j - 1) / 2), prime(j), (j - 1) / 2 < n→i(m) ⊦ countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) = [], j = 2, m = n, ~1 = (j - 1) / 2

Rule argument

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; 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%sub-zero-01: ⊦ i - 0 = i; from specification int-abs