Theorem 3sstr3g 4002

Description: Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 1-Oct-2000.)

Hypotheses

Ref	Expression
3sstr3g.1	⊢ (𝜑 → 𝐴 ⊆ 𝐵)
3sstr3g.2	⊢ 𝐴 = 𝐶
3sstr3g.3	⊢ 𝐵 = 𝐷

Assertion

Ref	Expression
3sstr3g	⊢ (𝜑 → 𝐶 ⊆ 𝐷)

Proof of Theorem 3sstr3g

Step	Hyp	Ref	Expression
1		3sstr3g.2	. . 3 ⊢ 𝐴 = 𝐶
2		3sstr3g.1	. . 3 ⊢ (𝜑 → 𝐴 ⊆ 𝐵)
3	1, 2	eqsstrrid 3989	. 2 ⊢ (𝜑 → 𝐶 ⊆ 𝐵)
4		3sstr3g.3	. 2 ⊢ 𝐵 = 𝐷
5	3, 4	sseqtrdi 3990	1 ⊢ (𝜑 → 𝐶 ⊆ 𝐷)

Syntax hints:   → wi 4   = wceq 1540   ⊆ wss 3917

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-9 2119  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-cleq 2722  df-ss 3934

This theorem is referenced by:  complss  4117  uniintsn  4952  fpwwe2lem12  10602  hmeocls  23662  hmeontr  23663  usgrumgruspgr  29116  chsscon3i  31397  pjss1coi  32099  mdslmd2i  32266  satffunlem2lem2  35400  ssbnd  37789  bnd2lem  37792  trclubgNEW  43614  nzss  44313