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Theorem ixpsnval 6798
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 6797 . 2 X𝑥 ∈ {𝑋}𝐵 = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵)}
2 ralsnsg 3672 . . . . 5 (𝑋𝑉 → (∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵[𝑋 / 𝑥](𝑓𝑥) ∈ 𝐵))
3 sbcel12g 3110 . . . . 5 (𝑋𝑉 → ([𝑋 / 𝑥](𝑓𝑥) ∈ 𝐵𝑋 / 𝑥(𝑓𝑥) ∈ 𝑋 / 𝑥𝐵))
4 csbfvg 5626 . . . . . 6 (𝑋𝑉𝑋 / 𝑥(𝑓𝑥) = (𝑓𝑋))
54eleq1d 2275 . . . . 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 2324 . 2 (𝑋𝑉 → {𝑓 ∣ (𝑓 Fn {𝑋} ∧ ∀𝑥 ∈ {𝑋} (𝑓𝑥) ∈ 𝐵)} = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)})
91, 8eqtrid 2251 1 (𝑋𝑉X𝑥 ∈ {𝑋}𝐵 = {𝑓 ∣ (𝑓 Fn {𝑋} ∧ (𝑓𝑋) ∈ 𝑋 / 𝑥𝐵)})
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104   = wceq 1373  wcel 2177  {cab 2192  wral 2485  [wsbc 3000  csb 3095  {csn 3635   Fn wfn 5272  cfv 5277  Xcixp 6795
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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188
This theorem depends on definitions:  df-bi 117  df-3an 983  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-ral 2490  df-rex 2491  df-v 2775  df-sbc 3001  df-csb 3096  df-un 3172  df-sn 3641  df-pr 3642  df-op 3644  df-uni 3854  df-br 4049  df-iota 5238  df-fn 5280  df-fv 5285  df-ixp 6796
This theorem is referenced by:  ixpsnbasval  14278
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