Theorem releqd 4802

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 4800	. 2 ⊢ (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵))
3	1, 2	syl 14	1 ⊢ (𝜑 → (Rel 𝐴 ↔ Rel 𝐵))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1395  Rel wrel 4723

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 3203  df-ss 3210  df-rel 4725

This theorem is referenced by:  dftpos3  6406  tposfo2  6411  tposf12  6413  imasaddfnlemg  13342  releqgg  13752  dvdsrd  14052  isunitd  14064  lmreltop  14861  cnprcl2k  14874