Theorem nfreu1 3110

Description: The setvar 𝑥 is not free in ∃!𝑥 ∈ 𝐴𝜑. (Contributed by NM, 19-Mar-1997.)

Assertion

Ref	Expression
nfreu1	⊢ Ⅎ𝑥∃!𝑥 ∈ 𝐴 𝜑

Proof of Theorem nfreu1

Step	Hyp	Ref	Expression
1		df-reu 2919	. 2 ⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
2		nfeu1 2480	. 2 ⊢ Ⅎ𝑥∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑)
3	1, 2	nfxfr 1779	1 ⊢ Ⅎ𝑥∃!𝑥 ∈ 𝐴 𝜑

Syntax hints:   ∧ wa 384  Ⅎwnf 1708   ∈ wcel 1990  ∃!weu 2470  ∃!wreu 2914

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-11 2034  ax-12 2047

This theorem depends on definitions:  df-bi 197  df-or 385  df-ex 1705  df-nf 1710  df-eu 2474  df-reu 2919

This theorem is referenced by:  riota2df  6631  2reu8  41192  iccpartdisj  41373