Theorem nfreu1 3396

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 3369	. 2 ⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
2		nfeu1 2617	. 2 ⊢ Ⅎ𝑥∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑)
3	1, 2	nfxfr 1885	1 ⊢ Ⅎ𝑥∃!𝑥 ∈ 𝐴 𝜑

This proof depends on syntax axioms:   ∧ wa 401  Ⅎwnf 1815   ∈ wcel 2146  ∃!weu 2596  ∃!wreu 3366

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-10 2179  ax-11 2196

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-mo 2567  df-eu 2597  df-reu 3369

This theorem is used by:  riota2df  7407  2reu8  48322  iccpartdisj  48658