Theorem 3sstr3g 3990

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 3977	. 2 ⊢ (𝜑 → 𝐶 ⊆ 𝐵)
4		3sstr3g.3	. 2 ⊢ 𝐵 = 𝐷
5	3, 4	sseqtrdi 3978	1 ⊢ (𝜑 → 𝐶 ⊆ 𝐷)

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ⊆ wss 3906

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2757  df-ss 3923

This theorem is used by:  complss  4105  uniintsn  4952  fpwwe2lem12  10644  hmeocls  23978  hmeontr  23979  usgrumgruspgr  29592  chsscon3i  31886  pjss1coi  32588  mdslmd2i  32755  satffunlem2lem2  35937  ssbnd  38499  bnd2lem  38502  trclubgNEW  44404  nzss  45087