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Proof step 11 from proof diff-primes-dups (previous step)

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 < 2 * abs(1 + (j - 1) / 2) ∧ (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)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: Abs-pos seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [1 + (j - 1) / 2] flags: rewrite

Additional information

Used simplifier rules: numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%div-ls-mone: 0 < j ⊦ i / (* j) < ~1 ↔ i + 2 * j ≤ 0; from specification int-div%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%numle-03: ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-le-03: ⊦ j ≤ i ∧ i < 0 → j < 0; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs Used terms: 1 + (j - 1) / 2