Theorem releqd 5655

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 5653	. 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 5562

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 2795

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 2802  df-cleq 2816  df-clel 2895  df-in 3945  df-ss 3954  df-rel 5564

This theorem is referenced by:  dftpos3  7912  tposfo2  7917  tposf12  7919  relexp0rel  14398  relexprelg  14399  relexpaddg  14414  imasaddfnlem  16803  imasvscafn  16812  cicer  17078  joindmss  17619  meetdmss  17633  mattpostpos  21065  cnextrel  22673  perpln1  26498  perpln2  26499  relfae  31508  satfrel  32616  dibvalrel  38301  dicvalrelN  38323  diclspsn  38332  dihvalrel  38417  dih1  38424  dihmeetlem4preN  38444