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Theorem exsnrex 3536
Description: There is a set being the element of a singleton if and only if there is an element of the singleton. (Contributed by Alexander van der Vekens, 1-Jan-2018.)
Assertion
Ref Expression
exsnrex (∃𝑥 𝑀 = {𝑥} ↔ ∃𝑥𝑀 𝑀 = {𝑥})

Proof of Theorem exsnrex
StepHypRef Expression
1 vex 2663 . . . . . 6 𝑥 ∈ V
21snid 3526 . . . . 5 𝑥 ∈ {𝑥}
3 eleq2 2181 . . . . 5 (𝑀 = {𝑥} → (𝑥𝑀𝑥 ∈ {𝑥}))
42, 3mpbiri 167 . . . 4 (𝑀 = {𝑥} → 𝑥𝑀)
54pm4.71ri 389 . . 3 (𝑀 = {𝑥} ↔ (𝑥𝑀𝑀 = {𝑥}))
65exbii 1569 . 2 (∃𝑥 𝑀 = {𝑥} ↔ ∃𝑥(𝑥𝑀𝑀 = {𝑥}))
7 df-rex 2399 . 2 (∃𝑥𝑀 𝑀 = {𝑥} ↔ ∃𝑥(𝑥𝑀𝑀 = {𝑥}))
86, 7bitr4i 186 1 (∃𝑥 𝑀 = {𝑥} ↔ ∃𝑥𝑀 𝑀 = {𝑥})
Colors of variables: wff set class
Syntax hints:  wa 103  wb 104   = wceq 1316  wex 1453  wcel 1465  wrex 2394  {csn 3497
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 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-bndl 1471  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500  ax-ext 2099
This theorem depends on definitions:  df-bi 116  df-tru 1319  df-nf 1422  df-sb 1721  df-clab 2104  df-cleq 2110  df-clel 2113  df-nfc 2247  df-rex 2399  df-v 2662  df-sn 3503
This theorem is referenced by: (None)
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