Theorem releq 4800

Description: Equality theorem for the relation predicate. (Contributed by NM, 1-Aug-1994.)

Assertion

Ref	Expression
releq	⊢ (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵))

Proof of Theorem releq

Step	Hyp	Ref	Expression
1		sseq1 3247	. 2 ⊢ (𝐴 = 𝐵 → (𝐴 ⊆ (V × V) ↔ 𝐵 ⊆ (V × V)))
2		df-rel 4725	. 2 ⊢ (Rel 𝐴 ↔ 𝐴 ⊆ (V × V))
3		df-rel 4725	. 2 ⊢ (Rel 𝐵 ↔ 𝐵 ⊆ (V × V))
4	1, 2, 3	3bitr4g 223	1 ⊢ (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1395  Vcvv 2799   ⊆ wss 3197   × cxp 4716  Rel wrel 4723

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-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-11 1552  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-in 3203  df-ss 3210  df-rel 4725

This theorem is referenced by:  releqi  4801  releqd  4802  dfrel2  5178  tposfn2  6410  ereq1  6685