Theorem sbcbr1g 5173

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 5172	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3875	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
3	2	breq2d 5126	. 2 ⊢ (𝐴 ∈ 𝑉 → (⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅𝐶))
4	1, 3	bitrd 282	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅𝐶))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∈ wcel 2146  [wsbc 3747  ⦋csb 3856   class class class wbr 5114

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-nfc 2915  df-rab 3420  df-v 3460  df-sbc 3748  df-csb 3857  df-dif 3911  df-un 3913  df-ss 3925  df-nul 4290  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-br 5115

This theorem is used by:  iscard4  44300  frege124d  44528  frege70  44700  frege92  44722  frege118  44748