Theorem sbcbr12g 3983

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 3982	. 2 ⊢ (𝐴 ∈ 𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3016	. . 3 ⊢ (𝐴 ∈ 𝐷 → ⦋𝐴 / 𝑥⦌𝑅 = 𝑅)
3	2	breqd 3940	. 2 ⊢ (𝐴 ∈ 𝐷 → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))
4	1, 3	bitrd 187	1 ⊢ (𝐴 ∈ 𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))

Syntax hints:   → wi 4   ↔ wb 104   ∈ wcel 1480  [wsbc 2909  ⦋csb 3003   class class class wbr 3929

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-3an 964  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-v 2688  df-sbc 2910  df-csb 3004  df-un 3075  df-sn 3533  df-pr 3534  df-op 3536  df-br 3930

This theorem is referenced by:  sbcbr1g  3984  sbcbr2g  3985