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Theorem fconstmpt 4685
Description: Representation of a constant function using the mapping operation. (Note that 𝑥 cannot appear free in 𝐵.) (Contributed by NM, 12-Oct-1999.) (Revised by Mario Carneiro, 16-Nov-2013.)
Assertion
Ref Expression
fconstmpt (𝐴 × {𝐵}) = (𝑥𝐴𝐵)
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem fconstmpt
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 velsn 3621 . . . 4 (𝑦 ∈ {𝐵} ↔ 𝑦 = 𝐵)
21anbi2i 457 . . 3 ((𝑥𝐴𝑦 ∈ {𝐵}) ↔ (𝑥𝐴𝑦 = 𝐵))
32opabbii 4082 . 2 {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 ∈ {𝐵})} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 = 𝐵)}
4 df-xp 4644 . 2 (𝐴 × {𝐵}) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 ∈ {𝐵})}
5 df-mpt 4078 . 2 (𝑥𝐴𝐵) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 = 𝐵)}
63, 4, 53eqtr4i 2218 1 (𝐴 × {𝐵}) = (𝑥𝐴𝐵)
Colors of variables: wff set class
Syntax hints:  wa 104   = wceq 1363  wcel 2158  {csn 3604  {copab 4075  cmpt 4076   × cxp 4636
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 710  ax-5 1457  ax-7 1458  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-8 1514  ax-10 1515  ax-11 1516  ax-i12 1517  ax-bndl 1519  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544  ax-i5r 1545  ax-ext 2169
This theorem depends on definitions:  df-bi 117  df-tru 1366  df-nf 1471  df-sb 1773  df-clab 2174  df-cleq 2180  df-clel 2183  df-nfc 2318  df-v 2751  df-sn 3610  df-opab 4077  df-mpt 4078  df-xp 4644
This theorem is referenced by:  fconst  5423  fcoconst  5700  fmptsn  5718  fconstmpo  5983  ofc12  6117  caofinvl  6119  xpexgALT  6148  inftonninf  10455  fser0const  10530  prod1dc  11608  cnmptc  14078  dvexp  14471  dvexp2  14472  dvmptidcn  14474  dvmptccn  14475  dvef  14484  nninfall  15055  nninfsellemeqinf  15062  exmidsbthrlem  15067
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