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Metamath Proof Explorer

Theorem hvsub0

Description: Subtraction of a zero vector. (Contributed by NM, 2-Apr-2000) (New usage is discouraged.)

Ref Expression
Assertion hvsub0 
|- ( A e. ~H -> ( A -h 0h ) = A )

Proof

Step Hyp Ref Expression
1 ax-hv0cl 
 |-  0h e. ~H
2 hvsubval 
 |-  ( ( A e. ~H /\ 0h e. ~H ) -> ( A -h 0h ) = ( A +h ( -u 1 .h 0h ) ) )
3 1 2 mpan2 
 |-  ( A e. ~H -> ( A -h 0h ) = ( A +h ( -u 1 .h 0h ) ) )
4 neg1cn 
 |-  -u 1 e. CC
5 hvmul0 
 |-  ( -u 1 e. CC -> ( -u 1 .h 0h ) = 0h )
6 4 5 ax-mp 
 |-  ( -u 1 .h 0h ) = 0h
7 6 oveq2i 
 |-  ( A +h ( -u 1 .h 0h ) ) = ( A +h 0h )
8 3 7 eqtrdi 
 |-  ( A e. ~H -> ( A -h 0h ) = ( A +h 0h ) )
9 ax-hvaddid 
 |-  ( A e. ~H -> ( A +h 0h ) = A )
10 8 9 eqtrd 
 |-  ( A e. ~H -> ( A -h 0h ) = A )