Theorem nfreu1 2723

Description: 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 2535	. 2 ⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
2		nfeu1 2097	. 2 ⊢ Ⅎ𝑥∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑)
3	1, 2	nfxfr 1527	1 ⊢ Ⅎ𝑥∃!𝑥 ∈ 𝐴 𝜑

This proof depends on syntax axioms:   ∧ wa 104  Ⅎwnf 1513  ∃!weu 2086   ∈ wcel 2209  ∃!wreu 2530

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-4 1563  ax-ial 1587

This proof depends on definitions:  df-bi 117  df-nf 1514  df-eu 2089  df-reu 2535

This theorem is used by:  riota2df  6060