Theorem releq 4808

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 3250	. 2 ⊢ (𝐴 = 𝐵 → (𝐴 ⊆ (V × V) ↔ 𝐵 ⊆ (V × V)))
2		df-rel 4732	. 2 ⊢ (Rel 𝐴 ↔ 𝐴 ⊆ (V × V))
3		df-rel 4732	. 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 1397  Vcvv 2802   ⊆ wss 3200   × cxp 4723  Rel wrel 4730

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 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-11 1554  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-in 3206  df-ss 3213  df-rel 4732

This theorem is referenced by:  releqi  4809  releqd  4810  dfrel2  5187  tposfn2  6431  ereq1  6708