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Theorem exsnrex 3505
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 2636 . . . . . 6 𝑥 ∈ V
21snid 3495 . . . . 5 𝑥 ∈ {𝑥}
3 eleq2 2158 . . . . 5 (𝑀 = {𝑥} → (𝑥𝑀𝑥 ∈ {𝑥}))
42, 3mpbiri 167 . . . 4 (𝑀 = {𝑥} → 𝑥𝑀)
54pm4.71ri 385 . . 3 (𝑀 = {𝑥} ↔ (𝑥𝑀𝑀 = {𝑥}))
65exbii 1548 . 2 (∃𝑥 𝑀 = {𝑥} ↔ ∃𝑥(𝑥𝑀𝑀 = {𝑥}))
7 df-rex 2376 . 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 1296  wex 1433  wcel 1445  wrex 2371  {csn 3466
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077
This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-rex 2376  df-v 2635  df-sn 3472
This theorem is referenced by: (None)
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