Theorem nfreu1 2669

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 2482	. 2 ⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
2		nfeu1 2056	. 2 ⊢ Ⅎ𝑥∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑)
3	1, 2	nfxfr 1488	1 ⊢ Ⅎ𝑥∃!𝑥 ∈ 𝐴 𝜑

Syntax hints:   ∧ wa 104  Ⅎwnf 1474  ∃!weu 2045   ∈ wcel 2167  ∃!wreu 2477

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-4 1524  ax-ial 1548

This theorem depends on definitions:  df-bi 117  df-nf 1475  df-eu 2048  df-reu 2482

This theorem is referenced by:  riota2df  5898