Theorem sbcbr1g 5156

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 5155	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3871	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
3	2	breq2d 5111	. 2 ⊢ (𝐴 ∈ 𝑉 → (⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅𝐶))
4	1, 3	bitrd 281	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅𝐶))

Syntax hints:   → wi 4   ↔ wb 208   ∈ wcel 2141  [wsbc 3744  ⦋csb 3852   class class class wbr 5099

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-10 2174  ax-11 2190  ax-12 2211  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1099  df-tru 1562  df-fal 1572  df-ex 1799  df-nf 1803  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-rab 3414  df-v 3455  df-sbc 3745  df-csb 3853  df-dif 3907  df-un 3909  df-ss 3921  df-nul 4286  df-if 4480  df-sn 4582  df-pr 4584  df-op 4588  df-br 5100

This theorem is referenced by:  iscard4  44073  frege124d  44301  frege70  44473  frege92  44495  frege118  44521