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Theorem ixpsnval 6865
Description: The value of an infinite Cartesian product with a singleton. (Contributed by AV, 3-Dec-2018.)
Assertion
Ref Expression
ixpsnval (𝑋𝑉X𝑥 ∈ {𝑋}𝐵 = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)})
Distinct variable groups:   𝐵,𝑓   𝑓,𝑉   𝑓,𝑋,𝑥
Allowed substitution hints:   𝐵(𝑥)   𝑉(𝑥)

Proof of Theorem ixpsnval
StepHypRef Expression
1 dfixp 6864 . 2 X𝑥 ∈ {𝑋}𝐵 = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵)}
2 ralsnsg 3704 . . . . 5 (𝑋𝑉 → (∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵[𝑋 / 𝑥](𝑓𝑥) ∈ 𝐵))
3 sbcel12g 3140 . . . . 5 (𝑋𝑉 → ([𝑋 / 𝑥](𝑓𝑥) ∈ 𝐵𝑋 / 𝑥(𝑓𝑥) ∈ 𝑋 / 𝑥𝐵))
4 csbfvg 5677 . . . . . 6 (𝑋𝑉𝑋 / 𝑥(𝑓𝑥) = (𝑓𝑋))
54eleq1d 2298 . . . . 5 (𝑋𝑉 → (𝑋 / 𝑥(𝑓𝑥) ∈ 𝑋 / 𝑥𝐵 ↔ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵))
62, 3, 53bitrd 214 . . . 4 (𝑋𝑉 → (∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵 ↔ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵))
76anbi2d 464 . . 3 (𝑋𝑉 → ((𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵) ↔ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)))
87abbidv 2347 . 2 (𝑋𝑉 → {𝑓 ∣ (𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵)} = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)})
91, 8eqtrid 2274 1 (𝑋𝑉X𝑥 ∈ {𝑋}𝐵 = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)})
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104   = wceq 1395  wcel 2200  {cab 2215  wral 2508  [wsbc 3029  csb 3125  {csn 3667   Fn wfn 5319  cfv 5324  Xcixp 6862
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211
This theorem depends on definitions:  df-bi 117  df-3an 1004  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ral 2513  df-rex 2514  df-v 2802  df-sbc 3030  df-csb 3126  df-un 3202  df-sn 3673  df-pr 3674  df-op 3676  df-uni 3892  df-br 4087  df-iota 5284  df-fn 5327  df-fv 5332  df-ixp 6863
This theorem is referenced by:  ixpsnbasval  14473
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