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Theorem exsnrex 3569
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 2690 . . . . . 6 𝑥 ∈ V
21snid 3559 . . . . 5 𝑥 ∈ {𝑥}
3 eleq2 2204 . . . . 5 (𝑀 = {𝑥} → (𝑥𝑀𝑥 ∈ {𝑥}))
42, 3mpbiri 167 . . . 4 (𝑀 = {𝑥} → 𝑥𝑀)
54pm4.71ri 390 . . 3 (𝑀 = {𝑥} ↔ (𝑥𝑀𝑀 = {𝑥}))
65exbii 1585 . 2 (∃𝑥 𝑀 = {𝑥} ↔ ∃𝑥(𝑥𝑀𝑀 = {𝑥}))
7 df-rex 2423 . 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 1332  wex 1469  wcel 1481  wrex 2418  {csn 3528
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 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122
This theorem depends on definitions:  df-bi 116  df-tru 1335  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-rex 2423  df-v 2689  df-sn 3534
This theorem is referenced by: (None)
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