Theorem 3sstr3g 3960

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 3947	. 2 ⊢ (𝜑 → 𝐶 ⊆ 𝐵)
4		3sstr3g.3	. 2 ⊢ 𝐵 = 𝐷
5	3, 4	sseqtrdi 3948	1 ⊢ (𝜑 → 𝐶 ⊆ 𝐷)

This proof depends on syntax axioms:   → wi 4   = wceq 1569   ⊆ wss 3895

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-cleq 2752  df-ss 3896

This theorem is used by:  complss  4087  uniintsn  4944  fpwwe2lem12  10727  hmeocls  24087  hmeontr  24088  usgrumgruspgr  29763  chsscon3i  32063  pjss1coi  32765  mdslmd2i  32932  satffunlem2lem2  36171  ssbnd  38723  bnd2lem  38726  trclubgNEW  44618  nzss  45301  cocanss1  45924