Theorem releqd 4808

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

Hypothesis

Ref	Expression
releqd.1	|- ( ph -> A = B )

Assertion

Ref	Expression
releqd	|- ( ph -> ( Rel A <-> Rel B ) )

Proof of Theorem releqd

Step	Hyp	Ref	Expression
1		releqd.1	. 2 |- ( ph -> A = B )
2		releq 4806	. 2 |- ( A = B -> ( Rel A <-> Rel B ) )
3	1, 2	syl 14	1 |- ( ph -> ( Rel A <-> Rel B ) )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1395   Rel wrel 4728

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 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-11 1552  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-in 3204  df-ss 3211  df-rel 4730

This theorem is referenced by:  dftpos3  6423  tposfo2  6428  tposf12  6430  imasaddfnlemg  13387  releqgg  13797  dvdsrd  14098  isunitd  14110  cnprcl2k  14920