Theorem releqd 5126

Description: Equality deduction for the relation predicate. (Contributed by NM, 8-Mar-2014.)

Hypothesis

Ref	Expression
releqd.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
releqd	⊢ (𝜑 → (Rel 𝐴 ↔ Rel 𝐵))

Proof of Theorem releqd

Step	Hyp	Ref	Expression
1		releqd.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		releq 5124	. 2 ⊢ (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵))
3	1, 2	syl 17	1 ⊢ (𝜑 → (Rel 𝐴 ↔ Rel 𝐵))

Syntax hints:   → wi 4   ↔ wb 195   = wceq 1475  Rel wrel 5043

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-in 3547  df-ss 3554  df-rel 5045

This theorem is referenced by:  dftpos3  7257  tposfo2  7262  tposf12  7264  relexp0rel  13625  relexprelg  13626  relexpaddg  13641  imasaddfnlem  16011  imasvscafn  16020  cicer  16289  joindmss  16830  meetdmss  16844  mattpostpos  20079  cnextrel  21677  perpln1  25405  perpln2  25406  relfae  29637  dibvalrel  35470  dicvalrelN  35492  diclspsn  35501  dihvalrel  35586  dih1  35593  dihmeetlem4preN  35613