Theorem releqd 5679

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 5677	. 2 ⊢ (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵))
3	1, 2	syl 17	1 ⊢ (𝜑 → (Rel 𝐴 ↔ Rel 𝐵))

Syntax hints:   → wi 4   ↔ wb 205   = wceq 1539  Rel wrel 5585

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-v 3424  df-in 3890  df-ss 3900  df-rel 5587

This theorem is referenced by:  dftpos3  8031  tposfo2  8036  tposf12  8038  relexp0rel  14676  relexprelg  14677  relexpreld  14679  relexpaddg  14692  imasaddfnlem  17156  imasvscafn  17165  cicer  17435  joindmss  18012  meetdmss  18026  mattpostpos  21511  cnextrel  23122  perpln1  26975  perpln2  26976  relfae  32115  satfrel  33229  dibvalrel  39104  dicvalrelN  39126  diclspsn  39135  dihvalrel  39220  dih1  39227  dihmeetlem4preN  39247