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Theorem ixpsnval 6525
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 6524 . 2 X𝑥 ∈ {𝑋}𝐵 = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵)}
2 ralsnsg 3508 . . . . 5 (𝑋𝑉 → (∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵[𝑋 / 𝑥](𝑓𝑥) ∈ 𝐵))
3 sbcel12g 2968 . . . . 5 (𝑋𝑉 → ([𝑋 / 𝑥](𝑓𝑥) ∈ 𝐵𝑋 / 𝑥(𝑓𝑥) ∈ 𝑋 / 𝑥𝐵))
4 csbfvg 5391 . . . . . 6 (𝑋𝑉𝑋 / 𝑥(𝑓𝑥) = (𝑓𝑋))
54eleq1d 2168 . . . . 5 (𝑋𝑉 → (𝑋 / 𝑥(𝑓𝑥) ∈ 𝑋 / 𝑥𝐵 ↔ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵))
62, 3, 53bitrd 213 . . . 4 (𝑋𝑉 → (∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵 ↔ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵))
76anbi2d 455 . . 3 (𝑋𝑉 → ((𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵) ↔ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)))
87abbidv 2217 . 2 (𝑋𝑉 → {𝑓 ∣ (𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵)} = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)})
91, 8syl5eq 2144 1 (𝑋𝑉X𝑥 ∈ {𝑋}𝐵 = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)})
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103   = wceq 1299  wcel 1448  {cab 2086  wral 2375  [wsbc 2862  csb 2955  {csn 3474   Fn wfn 5054  cfv 5059  Xcixp 6522
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 671  ax-5 1391  ax-7 1392  ax-gen 1393  ax-ie1 1437  ax-ie2 1438  ax-8 1450  ax-10 1451  ax-11 1452  ax-i12 1453  ax-bndl 1454  ax-4 1455  ax-17 1474  ax-i9 1478  ax-ial 1482  ax-i5r 1483  ax-ext 2082
This theorem depends on definitions:  df-bi 116  df-3an 932  df-tru 1302  df-nf 1405  df-sb 1704  df-clab 2087  df-cleq 2093  df-clel 2096  df-nfc 2229  df-ral 2380  df-rex 2381  df-v 2643  df-sbc 2863  df-csb 2956  df-un 3025  df-sn 3480  df-pr 3481  df-op 3483  df-uni 3684  df-br 3876  df-iota 5024  df-fn 5062  df-fv 5067  df-ixp 6523
This theorem is referenced by: (None)
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