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Theorem dfmpt3 5049
Description: Alternate definition for the "maps to" notation df-mpt 3848. (Contributed by Mario Carneiro, 30-Dec-2016.)
Assertion
Ref Expression
dfmpt3 (𝑥𝐴𝐵) = 𝑥𝐴 ({𝑥} × {𝐵})

Proof of Theorem dfmpt3
Dummy variables 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 df-mpt 3848 . 2 (𝑥𝐴𝐵) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 = 𝐵)}
2 velsn 3420 . . . . . . 7 (𝑦 ∈ {𝐵} ↔ 𝑦 = 𝐵)
32anbi2i 438 . . . . . 6 ((𝑥𝐴𝑦 ∈ {𝐵}) ↔ (𝑥𝐴𝑦 = 𝐵))
43anbi2i 438 . . . . 5 ((𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥𝐴𝑦 ∈ {𝐵})) ↔ (𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥𝐴𝑦 = 𝐵)))
542exbii 1513 . . . 4 (∃𝑥𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥𝐴𝑦 ∈ {𝐵})) ↔ ∃𝑥𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥𝐴𝑦 = 𝐵)))
6 eliunxp 4503 . . . 4 (𝑧 𝑥𝐴 ({𝑥} × {𝐵}) ↔ ∃𝑥𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥𝐴𝑦 ∈ {𝐵})))
7 elopab 4023 . . . 4 (𝑧 ∈ {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 = 𝐵)} ↔ ∃𝑥𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ (𝑥𝐴𝑦 = 𝐵)))
85, 6, 73bitr4i 205 . . 3 (𝑧 𝑥𝐴 ({𝑥} × {𝐵}) ↔ 𝑧 ∈ {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 = 𝐵)})
98eqriv 2053 . 2 𝑥𝐴 ({𝑥} × {𝐵}) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐴𝑦 = 𝐵)}
101, 9eqtr4i 2079 1 (𝑥𝐴𝐵) = 𝑥𝐴 ({𝑥} × {𝐵})
Colors of variables: wff set class
Syntax hints:  wa 101   = wceq 1259  wex 1397  wcel 1409  {csn 3403  cop 3406   ciun 3685  {copab 3845  cmpt 3846   × cxp 4371
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-14 1421  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038  ax-sep 3903  ax-pow 3955  ax-pr 3972
This theorem depends on definitions:  df-bi 114  df-3an 898  df-tru 1262  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-ral 2328  df-rex 2329  df-v 2576  df-sbc 2788  df-csb 2881  df-un 2950  df-in 2952  df-ss 2959  df-pw 3389  df-sn 3409  df-pr 3410  df-op 3412  df-iun 3687  df-opab 3847  df-mpt 3848  df-xp 4379  df-rel 4380
This theorem is referenced by:  dfmpt  5368  dfmptg  5370
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