Theorem sbcbr1g 5167

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

Assertion

Ref	Expression
sbcbr1g	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅𝐶))

Distinct variable groups:   𝑥,𝐶   𝑥,𝑅

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

Proof of Theorem sbcbr1g

Step	Hyp	Ref	Expression
1		sbcbr12g 5166	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3884	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
3	2	breq2d 5122	. 2 ⊢ (𝐴 ∈ 𝑉 → (⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅𝐶))
4	1, 3	bitrd 279	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅𝐶))

Syntax hints:   → wi 4   ↔ wb 206   ∈ wcel 2109  [wsbc 3756  ⦋csb 3865   class class class wbr 5110

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-nfc 2879  df-rab 3409  df-v 3452  df-sbc 3757  df-csb 3866  df-dif 3920  df-un 3922  df-ss 3934  df-nul 4300  df-if 4492  df-sn 4593  df-pr 4595  df-op 4599  df-br 5111

This theorem is referenced by:  iscard4  43529  frege124d  43757  frege70  43929  frege92  43951  frege118  43977