Theorem sseq12 3940

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 3938	. 2 ⊢ (𝐴 = 𝐵 → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶))
2		sseq2 3939	. 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 1570   ⊆ wss 3900

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2755  df-ss 3901

This theorem is used by:  sseq12i  3943  sorpsscmpl  7777  funcnvuni  7971  fiunlem  7981  sornom  10385  axdc3lem2  10559  ipole  18740  ipodrsima  18747  metsscmetcld  25669  funpsstri  36722  brredunds  39845  ismrcd2  43914  ismrc  43916