Metamath Proof Explorer


Theorem lspprcl

Description: The span of a pair is a subspace (frequently used special case of lspcl ). (Contributed by NM, 11-Apr-2015)

Ref Expression
Hypotheses lspval.v
|- V = ( Base ` W )
lspval.s
|- S = ( LSubSp ` W )
lspval.n
|- N = ( LSpan ` W )
lspprcl.w
|- ( ph -> W e. LMod )
lspprcl.x
|- ( ph -> X e. V )
lspprcl.y
|- ( ph -> Y e. V )
Assertion lspprcl
|- ( ph -> ( N ` { X , Y } ) e. S )

Proof

Step Hyp Ref Expression
1 lspval.v
 |-  V = ( Base ` W )
2 lspval.s
 |-  S = ( LSubSp ` W )
3 lspval.n
 |-  N = ( LSpan ` W )
4 lspprcl.w
 |-  ( ph -> W e. LMod )
5 lspprcl.x
 |-  ( ph -> X e. V )
6 lspprcl.y
 |-  ( ph -> Y e. V )
7 5 6 prssd
 |-  ( ph -> { X , Y } C_ V )
8 1 2 3 lspcl
 |-  ( ( W e. LMod /\ { X , Y } C_ V ) -> ( N ` { X , Y } ) e. S )
9 4 7 8 syl2anc
 |-  ( ph -> ( N ` { X , Y } ) e. S )