Theorem nfreu1 2525

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 2355	. 2 ⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
2		nfeu1 1952	. 2 ⊢ Ⅎ𝑥∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑)
3	1, 2	nfxfr 1403	1 ⊢ Ⅎ𝑥∃!𝑥 ∈ 𝐴 𝜑

Syntax hints:   ∧ wa 102  Ⅎwnf 1389   ∈ wcel 1433  ∃!weu 1941  ∃!wreu 2350

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-4 1440  ax-ial 1467

This theorem depends on definitions:  df-bi 115  df-nf 1390  df-eu 1944  df-reu 2355

This theorem is referenced by:  riota2df  5508