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Metamath Proof Explorer

Theorem ply1scln0

Description: Nonzero scalars create nonzero polynomials. (Contributed by Stefan O'Rear, 29-Mar-2015)

Ref Expression
Hypotheses ply1scl.p 
|- P = ( Poly1 ` R )
ply1scl.a 
|- A = ( algSc ` P )
ply1scl0.z 
|- .0. = ( 0g ` R )
ply1scl0.y 
|- Y = ( 0g ` P )
ply1scln0.k 
|- K = ( Base ` R )
Assertion ply1scln0 
|- ( ( R e. Ring /\ X e. K /\ X =/= .0. ) -> ( A ` X ) =/= Y )

Proof

Step Hyp Ref Expression
1 ply1scl.p 
 |-  P = ( Poly1 ` R )
2 ply1scl.a 
 |-  A = ( algSc ` P )
3 ply1scl0.z 
 |-  .0. = ( 0g ` R )
4 ply1scl0.y 
 |-  Y = ( 0g ` P )
5 ply1scln0.k 
 |-  K = ( Base ` R )
6 eqid 
 |-  ( Base ` P ) = ( Base ` P )
7 1 2 5 6 ply1sclf1 
 |-  ( R e. Ring -> A : K -1-1-> ( Base ` P ) )
8 7 adantr 
 |-  ( ( R e. Ring /\ X e. K ) -> A : K -1-1-> ( Base ` P ) )
9 simpr 
 |-  ( ( R e. Ring /\ X e. K ) -> X e. K )
10 5 3 ring0cl 
 |-  ( R e. Ring -> .0. e. K )
11 10 adantr 
 |-  ( ( R e. Ring /\ X e. K ) -> .0. e. K )
12 f1fveq 
 |-  ( ( A : K -1-1-> ( Base ` P ) /\ ( X e. K /\ .0. e. K ) ) -> ( ( A ` X ) = ( A ` .0. ) <-> X = .0. ) )
13 8 9 11 12 syl12anc 
 |-  ( ( R e. Ring /\ X e. K ) -> ( ( A ` X ) = ( A ` .0. ) <-> X = .0. ) )
14 13 biimpd 
 |-  ( ( R e. Ring /\ X e. K ) -> ( ( A ` X ) = ( A ` .0. ) -> X = .0. ) )
15 14 necon3d 
 |-  ( ( R e. Ring /\ X e. K ) -> ( X =/= .0. -> ( A ` X ) =/= ( A ` .0. ) ) )
16 15 3impia 
 |-  ( ( R e. Ring /\ X e. K /\ X =/= .0. ) -> ( A ` X ) =/= ( A ` .0. ) )
17 1 2 3 4 ply1scl0 
 |-  ( R e. Ring -> ( A ` .0. ) = Y )
18 17 3ad2ant1 
 |-  ( ( R e. Ring /\ X e. K /\ X =/= .0. ) -> ( A ` .0. ) = Y )
19 16 18 neeqtrd 
 |-  ( ( R e. Ring /\ X e. K /\ X =/= .0. ) -> ( A ` X ) =/= Y )