Theorem 3sstr3g 3965

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 3952	. 2 ⊢ (𝜑 → 𝐶 ⊆ 𝐵)
4		3sstr3g.3	. 2 ⊢ 𝐵 = 𝐷
5	3, 4	sseqtrdi 3953	1 ⊢ (𝜑 → 𝐶 ⊆ 𝐷)

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ⊆ wss 3900

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 1945  ax-6 2002  ax-7 2043  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2755  df-ss 3901

This theorem is used by:  complss  4090  uniintsn  4948  fpwwe2lem12  10751  hmeocls  24114  hmeontr  24115  usgrumgruspgr  29790  chsscon3i  32090  pjss1coi  32792  mdslmd2i  32959  satffunlem2lem2  36337  ssbnd  38913  bnd2lem  38916  trclubgNEW  44808  nzss  45491  cocanss1  46114