Theorem nfreu1 3397

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 3370	. 2 ⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
2		nfeu1 2617	. 2 ⊢ Ⅎ𝑥∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑)
3	1, 2	nfxfr 1886	1 ⊢ Ⅎ𝑥∃!𝑥 ∈ 𝐴 𝜑

This proof depends on syntax axioms:   ∧ wa 401  Ⅎwnf 1816   ∈ wcel 2147  ∃!weu 2596  ∃!wreu 3367

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-10 2180  ax-11 2197

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-mo 2567  df-eu 2597  df-reu 3370

This theorem is used by:  riota2df  7431  2reu8  48456  iccpartdisj  48792