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Theorem ixpsnval 6599
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 6598 . 2 X𝑥 ∈ {𝑋}𝐵 = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵)}
2 ralsnsg 3564 . . . . 5 (𝑋𝑉 → (∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵[𝑋 / 𝑥](𝑓𝑥) ∈ 𝐵))
3 sbcel12g 3018 . . . . 5 (𝑋𝑉 → ([𝑋 / 𝑥](𝑓𝑥) ∈ 𝐵𝑋 / 𝑥(𝑓𝑥) ∈ 𝑋 / 𝑥𝐵))
4 csbfvg 5463 . . . . . 6 (𝑋𝑉𝑋 / 𝑥(𝑓𝑥) = (𝑓𝑋))
54eleq1d 2209 . . . . 5 (𝑋𝑉 → (𝑋 / 𝑥(𝑓𝑥) ∈ 𝑋 / 𝑥𝐵 ↔ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵))
62, 3, 53bitrd 213 . . . 4 (𝑋𝑉 → (∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵 ↔ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵))
76anbi2d 460 . . 3 (𝑋𝑉 → ((𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵) ↔ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)))
87abbidv 2258 . 2 (𝑋𝑉 → {𝑓 ∣ (𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵)} = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)})
91, 8syl5eq 2185 1 (𝑋𝑉X𝑥 ∈ {𝑋}𝐵 = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)})
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103   = wceq 1332  wcel 1481  {cab 2126  wral 2417  [wsbc 2910  csb 3004  {csn 3528   Fn wfn 5122  cfv 5127  Xcixp 6596
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122
This theorem depends on definitions:  df-bi 116  df-3an 965  df-tru 1335  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-ral 2422  df-rex 2423  df-v 2689  df-sbc 2911  df-csb 3005  df-un 3076  df-sn 3534  df-pr 3535  df-op 3537  df-uni 3741  df-br 3934  df-iota 5092  df-fn 5130  df-fv 5135  df-ixp 6597
This theorem is referenced by: (None)
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