Theorem sbcbr12g 4103

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 4102	. 2 ⊢ (𝐴 ∈ 𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3108	. . 3 ⊢ (𝐴 ∈ 𝐷 → ⦋𝐴 / 𝑥⦌𝑅 = 𝑅)
3	2	breqd 4058	. 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 2177  [wsbc 2999  ⦋csb 3094   class class class wbr 4047

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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-3an 983  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-sbc 3000  df-csb 3095  df-un 3171  df-sn 3640  df-pr 3641  df-op 3643  df-br 4048

This theorem is referenced by:  sbcbr1g  4104  sbcbr2g  4105