Theorem sseq12 3127

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 3125	. 2 ⊢ (𝐴 = 𝐵 → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶))
2		sseq2 3126	. 2 ⊢ (𝐶 = 𝐷 → (𝐵 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷))
3	1, 2	sylan9bb 458	1 ⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104   = wceq 1332   ⊆ wss 3076

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-11 1485  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122

This theorem depends on definitions:  df-bi 116  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-in 3082  df-ss 3089

This theorem is referenced by:  sseq12i  3130  undifexmid  4125  exmidundif  4137  exmidundifim  4138  funcnvuni  5200  fun11iun  5396