Theorem nfreu1 3391

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 3360	. 2 ⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
2		nfeu1 2587	. 2 ⊢ Ⅎ𝑥∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑)
3	1, 2	nfxfr 1853	1 ⊢ Ⅎ𝑥∃!𝑥 ∈ 𝐴 𝜑

Syntax hints:   ∧ wa 395  Ⅎwnf 1783   ∈ wcel 2108  ∃!weu 2567  ∃!wreu 3357

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-10 2141  ax-11 2157  ax-12 2177

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-ex 1780  df-nf 1784  df-mo 2539  df-eu 2568  df-reu 3360

This theorem is referenced by:  riota2df  7383  2reu8  47089  iccpartdisj  47399