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Theorem fconstmpt 4387
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 3392 . . . 4 (𝑦 ∈ {𝐵} ↔ 𝑦 = 𝐵)
21anbi2i 430 . . 3 ((𝑥𝐴𝑦 ∈ {𝐵}) ↔ (𝑥𝐴𝑦 = 𝐵))
32opabbii 3824 . 2 {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 ∈ {𝐵})} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 = 𝐵)}
4 df-xp 4351 . 2 (𝐴 × {𝐵}) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 ∈ {𝐵})}
5 df-mpt 3820 . 2 (𝑥𝐴𝐵) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 = 𝐵)}
63, 4, 53eqtr4i 2070 1 (𝐴 × {𝐵}) = (𝑥𝐴𝐵)
Colors of variables: wff set class
Syntax hints:  wa 97   = wceq 1243  wcel 1393  {csn 3375  {copab 3817  cmpt 3818   × cxp 4343
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022
This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-sn 3381  df-opab 3819  df-mpt 3820  df-xp 4351
This theorem is referenced by:  fconst  5082  fcoconst  5334  fmptsn  5352  ofc12  5731  caofinvl  5733  xpexgALT  5760
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