Theorem sbcbr12g 4144

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 4143	. 2 ⊢ (𝐴 ∈ 𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3141	. . 3 ⊢ (𝐴 ∈ 𝐷 → ⦋𝐴 / 𝑥⦌𝑅 = 𝑅)
3	2	breqd 4099	. 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 2202  [wsbc 3031  ⦋csb 3127   class class class wbr 4088

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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-3an 1006  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-v 2804  df-sbc 3032  df-csb 3128  df-un 3204  df-sn 3675  df-pr 3676  df-op 3678  df-br 4089

This theorem is referenced by:  sbcbr1g  4145  sbcbr2g  4146