Theorem eqeq12 2365

Description: Equality relationship among 4 classes. (Contributed by NM, 3-Aug-1994.)

Assertion

Ref	Expression
eqeq12	⊢ ((A = B ∧ C = D) → (A = C ↔ B = D))

Proof of Theorem eqeq12

Step	Hyp	Ref	Expression
1		eqeq1 2359	. 2 ⊢ (A = B → (A = C ↔ B = C))
2		eqeq2 2362	. 2 ⊢ (C = D → (B = C ↔ B = D))
3	1, 2	sylan9bb 680	1 ⊢ ((A = B ∧ C = D) → (A = C ↔ B = D))

Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358   = wceq 1642

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-11 1746  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1542  df-cleq 2346

This theorem is referenced by:  eqeq12i  2366  eqeq12d  2367  eqeqan12d  2368  dfpw12  4301  fununiq  5517  fntxp  5804  pw1fnf1o  5855  fundmen  6043  ncdisjeq  6148  peano4nc  6150  sbth  6206  tc11  6227  fnfrec  6318