Theorem sbcbr2g 5149

Description: Move substitution in and out of a binary relation. (Contributed by NM, 13-Dec-2005.)

Assertion

Ref	Expression
sbcbr2g	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ 𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))

Distinct variable groups:   𝑥,𝐵   𝑥,𝑅

Allowed substitution hints:   𝐴(𝑥)   𝐶(𝑥)   𝑉(𝑥)

Proof of Theorem sbcbr2g

Step	Hyp	Ref	Expression
1		sbcbr12g 5147	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3869	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3	2	breq1d 5101	. 2 ⊢ (𝐴 ∈ 𝑉 → (⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))
4	1, 3	bitrd 279	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ 𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))

Syntax hints:   → wi 4   ↔ wb 206   ∈ wcel 2111  [wsbc 3741  ⦋csb 3850   class class class wbr 5091

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-10 2144  ax-11 2160  ax-12 2180  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-nf 1785  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-nfc 2881  df-rab 3396  df-v 3438  df-sbc 3742  df-csb 3851  df-dif 3905  df-un 3907  df-ss 3919  df-nul 4284  df-if 4476  df-sn 4577  df-pr 4579  df-op 4583  df-br 5092

This theorem is referenced by:  prmgaplem7  16966  telgsums  19903  fvmptnn04if  22762  bnj110  34865  frege124d  43793  frege72  43967  frege91  43986  frege116  44011  frege120  44015