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Proof step 30 from proof diff-primes-dups (previous step)

i ∈ countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j), 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)), 0 ≤ i ∧ i < j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 31:
⊦ i0 ∈ countFun(i, j, intfun) ↔ (∃ j0. i ≤ j0 ∧ j0 ≤ j ∧ intfun(j0) = i0)
Proof step 32:
∃ j0. 0 ≤ j0 ∧ j0 ≤ (j - 1) / 2 ∧ (~1 - j0 ^2) |%| j = i, 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)), 0 ≤ i ∧ i < j

Rule argument

spec: intlist inst: name: countFun-in seq: ⊦ i0 ∈ countFun(i, j, intfun) ↔ (∃ j0. i ≤ j0 ∧ j0 ≤ j ∧ intfun(j0) = i0) subst: [intfun, j, i, i0] → [λ i. (~1 - i ^2) |%| j, (j - 1) / 2, 0, i] flags: rewrite

Additional information

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