Theorem sseq12i 3198

Description: An equality inference for the subclass relationship. (Contributed by NM, 31-May-1999.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)

Hypotheses

Ref	Expression
sseq1i.1	⊢ 𝐴 = 𝐵
sseq12i.2	⊢ 𝐶 = 𝐷

Assertion

Ref	Expression
sseq12i	⊢ (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷)

Proof of Theorem sseq12i

Step	Hyp	Ref	Expression
1		sseq1i.1	. 2 ⊢ 𝐴 = 𝐵
2		sseq12i.2	. 2 ⊢ 𝐶 = 𝐷
3		sseq12 3195	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷))
4	1, 2, 3	mp2an 426	1 ⊢ (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷)

Syntax hints:   ↔ wb 105   = wceq 1364   ⊆ wss 3144

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 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-11 1517  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2171

This theorem depends on definitions:  df-bi 117  df-nf 1472  df-sb 1774  df-clab 2176  df-cleq 2182  df-clel 2185  df-in 3150  df-ss 3157

This theorem is referenced by:  3sstr3i  3210  3sstr4i  3211  3sstr3g  3212  3sstr4g  3213  ss2rab  3246