Theorem sseq12 3192

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 3190	. 2 ⊢ (𝐴 = 𝐵 → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶))
2		sseq2 3191	. 2 ⊢ (𝐶 = 𝐷 → (𝐵 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷))
3	1, 2	sylan9bb 462	1 ⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1363   ⊆ wss 3141

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 1457  ax-7 1458  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-8 1514  ax-11 1516  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544  ax-i5r 1545  ax-ext 2169

This theorem depends on definitions:  df-bi 117  df-nf 1471  df-sb 1773  df-clab 2174  df-cleq 2180  df-clel 2183  df-in 3147  df-ss 3154

This theorem is referenced by:  sseq12i  3195  undifexmid  4205  exmidundif  4218  exmidundifim  4219  funcnvuni  5297  fun11iun  5494