Theorem sseq12 3967

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 3965	. 2 ⊢ (𝐴 = 𝐵 → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶))
2		sseq2 3966	. 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 3908

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 1943  ax-6 2000  ax-7 2041  ax-9 2156  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2758  df-ss 3925

This theorem is used by:  sseq12i  3970  sorpsscmpl  7744  funcnvuni  7938  fiunlem  7948  sornom  10279  axdc3lem2  10453  ipole  18615  ipodrsima  18622  metsscmetcld  25511  funpsstri  36279  brredunds  39400  ismrcd2  43471  ismrc  43473