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Theorem issetf 3199
Description: A version of isset 3198 that does not require 𝑥 and 𝐴 to be distinct. (Contributed by Andrew Salmon, 6-Jun-2011.) (Revised by Mario Carneiro, 10-Oct-2016.)
Hypothesis
Ref Expression
issetf.1 𝑥𝐴
Assertion
Ref Expression
issetf (𝐴 ∈ V ↔ ∃𝑥 𝑥 = 𝐴)

Proof of Theorem issetf
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 isset 3198 . 2 (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
2 issetf.1 . . . 4 𝑥𝐴
32nfeq2 2782 . . 3 𝑥 𝑦 = 𝐴
4 nfv 1845 . . 3 𝑦 𝑥 = 𝐴
5 eqeq1 2630 . . 3 (𝑦 = 𝑥 → (𝑦 = 𝐴𝑥 = 𝐴))
63, 4, 5cbvex 2276 . 2 (∃𝑦 𝑦 = 𝐴 ↔ ∃𝑥 𝑥 = 𝐴)
71, 6bitri 264 1 (𝐴 ∈ V ↔ ∃𝑥 𝑥 = 𝐴)
Colors of variables: wff setvar class
Syntax hints:  wb 196   = wceq 1480  wex 1701  wcel 1992  wnfc 2754  Vcvv 3191
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1841  ax-6 1890  ax-7 1937  ax-9 2001  ax-10 2021  ax-11 2036  ax-12 2049  ax-13 2250  ax-ext 2606
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1883  df-clab 2613  df-cleq 2619  df-clel 2622  df-nfc 2756  df-v 3193
This theorem is referenced by:  vtoclgf  3255  spcimgft  3275
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