Theorem rabdif 4267

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 4227	. 2 ⊢ ({𝑥 ∣ 𝜑} ∩ (𝐴 ∖ 𝐵)) = (({𝑥 ∣ 𝜑} ∩ 𝐴) ∖ 𝐵)
2		dfrab2 4266	. 2 ⊢ {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑} = ({𝑥 ∣ 𝜑} ∩ (𝐴 ∖ 𝐵))
3		dfrab2 4266	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = ({𝑥 ∣ 𝜑} ∩ 𝐴)
4	3	difeq1i 4070	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ 𝐵) = (({𝑥 ∣ 𝜑} ∩ 𝐴) ∖ 𝐵)
5	1, 2, 4	3eqtr4ri 2795	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ 𝐵) = {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑}

This proof depends on syntax axioms:   = wceq 1570  {cab 2739  {crab 3413   ∖ cdif 3896   ∩ cin 3898

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-rab 3414  df-v 3453  df-dif 3902  df-in 3906

This theorem is used by:  prjspeclsp  43620  dfclnbgr4  48891  clnbupgr  48900  dfnbgr5  48918  dfnbgr6  48924