Theorem nfreu1 2637

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 2451	. 2 ⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
2		nfeu1 2025	. 2 ⊢ Ⅎ𝑥∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑)
3	1, 2	nfxfr 1462	1 ⊢ Ⅎ𝑥∃!𝑥 ∈ 𝐴 𝜑

Syntax hints:   ∧ wa 103  Ⅎwnf 1448  ∃!weu 2014   ∈ wcel 2136  ∃!wreu 2446

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-4 1498  ax-ial 1522

This theorem depends on definitions:  df-bi 116  df-nf 1449  df-eu 2017  df-reu 2451

This theorem is referenced by:  riota2df  5818