Theorem sseq12 3935

Description: Equality theorem for the subclass relationship. (Contributed by NM, 31-May-1999.)

Assertion

Ref	Expression
sseq12	⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷))

Proof of Theorem sseq12

Step	Hyp	Ref	Expression
1		sseq1 3933	. 2 ⊢ (𝐴 = 𝐵 → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶))
2		sseq2 3934	. 2 ⊢ (𝐶 = 𝐷 → (𝐵 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷))
3	1, 2	sylan9bb 519	1 ⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569   ⊆ wss 3895

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-cleq 2752  df-ss 3896

This theorem is used by:  sseq12i  3938  sorpsscmpl  7750  funcnvuni  7944  fiunlem  7954  sornom  10358  axdc3lem2  10532  ipole  18713  ipodrsima  18720  metsscmetcld  25641  funpsstri  36623  brredunds  39735  ismrcd2  43803  ismrc  43805