Theorem sseqtrd 2100

Description: Substitution of equality into a subclass relationship.

Hypotheses

Ref	Expression
sseqtrd.1	|- (ph -> A (_ B)
sseqtrd.2	|- (ph -> B = C)

Assertion

Ref	Expression
sseqtrd	|- (ph -> A (_ C)

Proof of Theorem sseqtrd

Step	Hyp	Ref	Expression
1		sseqtrd.1	. 2 |- (ph -> A (_ B)
2		sseqtrd.2	. . 3 |- (ph -> B = C)
3	2	sseq2d 2092	. 2 |- (ph -> (A (_ B <-> A (_ C))
4	1, 3	mpbid 195	1 |- (ph -> A (_ C)

Syntax hints:   -> wi 3   = wceq 958   (_ wss 2050

This theorem is referenced by:  sseqtr4d 2101  fconst4 3857  nnaword2 4251  unifiOLD 4570  r1val1 4668  rankr1id 4707  fodom 4808  tgclt 7623  tgss2t 7636  2basgent 7640  bastop2 7642  clsss2 7700  iscncl 7767  cnconst 7777  dnsconst 7785  unirnbl 7872  metelcls 7962  shsub2t 9284  ococint 9292  spanssoc 9314  shub2t 9348  chub2t 9426  spanpr 9498  ssmd1 10233  mdslj1 10241  mdslj2 10242  mdslmd3 10254  mdexch 10257  irredlem1 10312  atcvat3 10318  atcvat4 10319  mdsymlem1 10325  mdsymlem5 10329

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-in 2054  df-ss 2056

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