Theorem nfre1 2593

Description: 𝑥 is not free in ∃𝑥 ∈ 𝐴𝜑. (Contributed by NM, 19-Mar-1997.) (Revised by Mario Carneiro, 7-Oct-2016.)

Assertion

Ref	Expression
nfre1	⊢ Ⅎ𝑥∃𝑥 ∈ 𝐴 𝜑

Proof of Theorem nfre1

Step	Hyp	Ref	Expression
1		df-rex 2534	. 2 ⊢ (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
2		nfe1 1549	. 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  ∃wex 1545   ∈ wcel 2209  ∃wrex 2529

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-gen 1502  ax-ie1 1546

This proof depends on definitions:  df-bi 117  df-nf 1514  df-rex 2534

This theorem is used by:  r19.29an  2693  nfiu1  4042  fun11iun  5660  eusvobj2  6071  fodjuomnilemdc  7485  ismkvnex  7496  prarloclem3step  7864  prmuloc2  7935  ltexprlemm  7968  caucvgprprlemaddq  8076  caucvgsrlemgt1  8163  axpre-suploclemres  8269  supinfneg  10005  infsupneg  10006  lbzbi  10026  divalglemeunn  12707  divalglemeuneg  12709  bezoutlemmain  12794  bezout  12807  lss1d  14804  pw1nct  17199  isomninnlem  17245  trirec0  17260  ismkvnnlem  17269