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Theorem ixpsnval 6658
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 6657 . 2 X𝑥 ∈ {𝑋}𝐵 = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵)}
2 ralsnsg 3607 . . . . 5 (𝑋𝑉 → (∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵[𝑋 / 𝑥](𝑓𝑥) ∈ 𝐵))
3 sbcel12g 3055 . . . . 5 (𝑋𝑉 → ([𝑋 / 𝑥](𝑓𝑥) ∈ 𝐵𝑋 / 𝑥(𝑓𝑥) ∈ 𝑋 / 𝑥𝐵))
4 csbfvg 5518 . . . . . 6 (𝑋𝑉𝑋 / 𝑥(𝑓𝑥) = (𝑓𝑋))
54eleq1d 2233 . . . . 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 2282 . 2 (𝑋𝑉 → {𝑓 ∣ (𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵)} = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)})
91, 8syl5eq 2209 1 (𝑋𝑉X𝑥 ∈ {𝑋}𝐵 = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)})
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103   = wceq 1342  wcel 2135  {cab 2150  wral 2442  [wsbc 2946  csb 3040  {csn 3570   Fn wfn 5177  cfv 5182  Xcixp 6655
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 1434  ax-7 1435  ax-gen 1436  ax-ie1 1480  ax-ie2 1481  ax-8 1491  ax-10 1492  ax-11 1493  ax-i12 1494  ax-bndl 1496  ax-4 1497  ax-17 1513  ax-i9 1517  ax-ial 1521  ax-i5r 1522  ax-ext 2146
This theorem depends on definitions:  df-bi 116  df-3an 969  df-tru 1345  df-nf 1448  df-sb 1750  df-clab 2151  df-cleq 2157  df-clel 2160  df-nfc 2295  df-ral 2447  df-rex 2448  df-v 2723  df-sbc 2947  df-csb 3041  df-un 3115  df-sn 3576  df-pr 3577  df-op 3579  df-uni 3784  df-br 3977  df-iota 5147  df-fn 5185  df-fv 5190  df-ixp 6656
This theorem is referenced by: (None)
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