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Theorem nfreu1 2498
Description: 𝑥 is not free in ∃!𝑥𝐴𝜑. (Contributed by NM, 19-Mar-1997.)
Assertion
Ref Expression
nfreu1 𝑥∃!𝑥𝐴 𝜑

Proof of Theorem nfreu1
StepHypRef Expression
1 df-reu 2330 . 2 (∃!𝑥𝐴 𝜑 ↔ ∃!𝑥(𝑥𝐴𝜑))
2 nfeu1 1927 . 2 𝑥∃!𝑥(𝑥𝐴𝜑)
31, 2nfxfr 1379 1 𝑥∃!𝑥𝐴 𝜑
Colors of variables: wff set class
Syntax hints:  wa 101  wnf 1365  wcel 1409  ∃!weu 1916  ∃!wreu 2325
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-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-4 1416  ax-ial 1443
This theorem depends on definitions:  df-bi 114  df-nf 1366  df-eu 1919  df-reu 2330
This theorem is referenced by:  riota2df  5515
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