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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j)[m] = countFun(0, (j - 1) / 2, λ i. (~1 - 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. (~1 - i ^2) |%| j) = [], j = 2, m = n, ~1 = (j - 1) / 2

Rule argument

Additional information

Used simplifier rules: numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%abs-zero: ⊦ i→n(i) = 0 ↔ i = 0; from specification intnat%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; 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