Theorem sbcbr1g 4058

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

Assertion

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

Distinct variable groups:   𝑥,𝐶   𝑥,𝑅

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

Proof of Theorem sbcbr1g

Step	Hyp	Ref	Expression
1		sbcbr12g 4057	. 2 ⊢ (𝐴 ∈ 𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3071	. . 3 ⊢ (𝐴 ∈ 𝐷 → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
3	2	breq2d 4014	. 2 ⊢ (𝐴 ∈ 𝐷 → (⦋𝐴 / 𝑥⦌𝐵𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅𝐶))
4	1, 3	bitrd 188	1 ⊢ (𝐴 ∈ 𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵𝑅𝐶))

Syntax hints:   → wi 4   ↔ wb 105   ∈ wcel 2148  [wsbc 2962  ⦋csb 3057   class class class wbr 4002

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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-3an 980  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2739  df-sbc 2963  df-csb 3058  df-un 3133  df-sn 3598  df-pr 3599  df-op 3601  df-br 4003

This theorem is referenced by: (None)