Theorem sbcbr12g 4164

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

Assertion

Ref	Expression
sbcbr12g	⊢ (𝐴 ∈ 𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))

Distinct variable group:   𝑥,𝑅

Allowed substitution hints:   𝐴(𝑥)   𝐵(𝑥)   𝐶(𝑥)   𝐷(𝑥)

Proof of Theorem sbcbr12g

Step	Hyp	Ref	Expression
1		sbcbrg 4163	. 2 ⊢ (𝐴 ∈ 𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3151	. . 3 ⊢ (𝐴 ∈ 𝐷 → ⦋𝐴 / 𝑥⦌𝑅 = 𝑅)
3	2	breqd 4119	. 2 ⊢ (𝐴 ∈ 𝐷 → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))
4	1, 3	bitrd 188	1 ⊢ (𝐴 ∈ 𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))

Syntax hints:   → wi 4   ↔ wb 105   ∈ wcel 2203  [wsbc 3041  ⦋csb 3137   class class class wbr 4108

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2214

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228  df-nfc 2373  df-v 2814  df-sbc 3042  df-csb 3138  df-un 3214  df-sn 3694  df-pr 3695  df-op 3697  df-br 4109

This theorem is referenced by:  sbcbr1g  4165  sbcbr2g  4166