Theorem metss 9101

Description: If two metrics are in a subset relationship, so are their base sets.

Hypotheses

Ref	Expression
metss.1	|- X = dom dom C
metss.2	|- Y = dom dom D

Assertion

Ref	Expression
metss	|- (C C_ D -> X C_ Y)

Proof of Theorem metss

Step	Hyp	Ref	Expression
1		dmss 4156	. . 3 |- (C C_ D -> dom C C_ dom D)
2		dmss 4156	. . 3 |- (dom C C_ dom D -> dom dom C C_ dom dom D)
3	1, 2	syl 12	. 2 |- (C C_ D -> dom dom C C_ dom dom D)
4		metss.1	. 2 |- X = dom dom C
5		metss.2	. 2 |- Y = dom dom D
6	3, 4, 5	3sstr4g 2658	1 |- (C C_ D -> X C_ Y)

Syntax hints:   -> wi 3   = wceq 1298   C_ wss 2593  dom cdm 3986

This theorem is referenced by:  metcnss 9176  metcnss2 9177

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-8 1306  ax-9 1307  ax-10 1308  ax-11 1309  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580  ax-11o 1588  ax-ext 1865

This theorem depends on definitions:  df-bi 164  df-or 241  df-an 242  df-ex 1327  df-sb 1536  df-clab 1872  df-cleq 1877  df-clel 1880  df-v 2294  df-un 2600  df-in 2603  df-ss 2605  df-sn 3049  df-pr 3050  df-op 3053  df-br 3339  df-dm 4004

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