Theorem rabdif 4266

Description: Move difference in and out of a restricted class abstraction. (Contributed by Steven Nguyen, 6-Jun-2023.)

Assertion

Ref	Expression
rabdif	⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ 𝐵) = {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑}

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem rabdif

Step	Hyp	Ref	Expression
1		indif2 4226	. 2 ⊢ ({𝑥 ∣ 𝜑} ∩ (𝐴 ∖ 𝐵)) = (({𝑥 ∣ 𝜑} ∩ 𝐴) ∖ 𝐵)
2		dfrab2 4265	. 2 ⊢ {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑} = ({𝑥 ∣ 𝜑} ∩ (𝐴 ∖ 𝐵))
3		dfrab2 4265	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = ({𝑥 ∣ 𝜑} ∩ 𝐴)
4	3	difeq1i 4059	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ 𝐵) = (({𝑥 ∣ 𝜑} ∩ 𝐴) ∖ 𝐵)
5	1, 2, 4	3eqtr4ri 2794	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ 𝐵) = {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑}

This proof depends on syntax axioms:   = wceq 1569  {cab 2738  {crab 3412   ∖ cdif 4045   ∩ cin 4089

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-dif 4047  df-in 4091

This theorem is used by:  prjspeclsp  43734  dfclnbgr4  49016  clnbupgr  49025  dfnbgr5  49043  dfnbgr6  49049