Theorem releq 4680

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 3160	. 2 ⊢ (𝐴 = 𝐵 → (𝐴 ⊆ (V × V) ↔ 𝐵 ⊆ (V × V)))
2		df-rel 4605	. 2 ⊢ (Rel 𝐴 ↔ 𝐴 ⊆ (V × V))
3		df-rel 4605	. 2 ⊢ (Rel 𝐵 ↔ 𝐵 ⊆ (V × V))
4	1, 2, 3	3bitr4g 222	1 ⊢ (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵))

Syntax hints:   → wi 4   ↔ wb 104   = wceq 1342  Vcvv 2721   ⊆ wss 3111   × cxp 4596  Rel wrel 4603

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1434  ax-7 1435  ax-gen 1436  ax-ie1 1480  ax-ie2 1481  ax-8 1491  ax-11 1493  ax-4 1497  ax-17 1513  ax-i9 1517  ax-ial 1521  ax-i5r 1522  ax-ext 2146

This theorem depends on definitions:  df-bi 116  df-nf 1448  df-sb 1750  df-clab 2151  df-cleq 2157  df-clel 2160  df-in 3117  df-ss 3124  df-rel 4605

This theorem is referenced by:  releqi  4681  releqd  4682  dfrel2  5048  tposfn2  6225  ereq1  6499