Theorem releqd 4767

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 4765	. 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 1373  Rel wrel 4688

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 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-11 1530  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-in 3176  df-ss 3183  df-rel 4690

This theorem is referenced by:  dftpos3  6361  tposfo2  6366  tposf12  6368  imasaddfnlemg  13221  releqgg  13631  dvdsrd  13931  isunitd  13943  lmreltop  14740  cnprcl2k  14753