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 3870	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
3	2	breq2d 5118	. 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 2147  [wsbc 3742  ⦋csb 3851   class class class wbr 5106

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-rab 3417  df-v 3456  df-sbc 3743  df-csb 3852  df-ss 3901  df-dif 4050  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107

This theorem is used by:  iscard4  44737  frege124d  44964  frege70  45136  frege92  45158  frege118  45184