Theorem releqd 5653

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 5651	. 2 ⊢ (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵))
3	1, 2	syl 17	1 ⊢ (𝜑 → (Rel 𝐴 ↔ Rel 𝐵))

Syntax hints:   → wi 4   ↔ wb 208   = wceq 1537  Rel wrel 5560

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2800  df-cleq 2814  df-clel 2893  df-in 3943  df-ss 3952  df-rel 5562

This theorem is referenced by:  dftpos3  7910  tposfo2  7915  tposf12  7917  relexp0rel  14396  relexprelg  14397  relexpaddg  14412  imasaddfnlem  16801  imasvscafn  16810  cicer  17076  joindmss  17617  meetdmss  17631  mattpostpos  21063  cnextrel  22671  perpln1  26496  perpln2  26497  relfae  31506  satfrel  32614  dibvalrel  38314  dicvalrelN  38336  diclspsn  38345  dihvalrel  38430  dih1  38437  dihmeetlem4preN  38457