theorem
  p1+p2 = (p1.1+p2.1)*<e1>+(p1.2+p2.2)*<e2>+(p1.3+p2.3)*<e3>
proof
A1:  <e1>.1 = 1 & <e1>.2 = 0 & <e1>.3 = 0;
A2:  <e2>.1 = 0 & <e2>.2 = 1 & <e2>.3 = 0;
A3:  <e3>.1 = 0 & <e3>.2 = 0 & <e3>.3 = 1;
A4: (p1+p2).1 = p1.1+p2.1 by RVSUM_1:11;
A5: (p1+p2).2 = p1.2+p2.2 by RVSUM_1:11;
A6: (p1+p2).3 = p1.3+p2.3 by RVSUM_1:11;
A7: (p1+p2).1*<e1> + (p1+p2).2*<e2> + (p1+p2).3*<e3>
  = |[ (p1+p2).1*1,(p1+p2).1*0,(p1+p2).1*0 ]| + (p1+p2).2*<e2>
    + (p1+p2).3*<e3> by A1,Lm1
 .= |[ (p1+p2).1,0,0 ]| + |[ (p1+p2).2*0,(p1+p2).2*1,(p1+p2).2*0 ]|
    + (p1+p2).3*<e3> by A2,Lm1
 .= |[ (p1+p2).1,0,0 ]| + |[ 0,(p1+p2).2,0 ]|
    + |[ (p1+p2).3*0,(p1+p2).3*0,(p1+p2).3*1 ]| by A3,Lm1
 .= |[ (p1+p2).1,0,0 ]| + |[ 0,(p1+p2).2,0 ]| + |[ 0,0,(p1+p2).3 ]|;
A8: |[ (p1+p2).1,0,0 ]|.1 = (p1+p2).1 & |[ (p1+p2).1,0,0 ]|.2 = 0 &
    |[ (p1+p2).1,0,0 ]|.3 = 0;
A9: |[ 0,(p1+p2).2,0 ]|.1 = 0 & |[ 0,(p1+p2).2,0 ]|.2 = (p1+p2).2 &
    |[ 0,(p1+p2).2,0 ]|.3 = 0;
A10: |[ 0,0,(p1+p2).3 ]|.1 = 0 & |[ 0,0,(p1+p2).3 ]|.2 = 0 &
    |[ 0,0,(p1+p2).3 ]|.3 = (p1+p2).3;
A11: (p1+p2).1*<e1> + (p1+p2).2*<e2> + (p1+p2).3*<e3>
  = |[ (p1+p2).1+0,0+(p1+p2).2,0+0 ]| + |[ 0,0,(p1+p2).3 ]|
    by A7,A8,A9,Lm2
 .= |[ (p1+p2).1,(p1+p2).2,0 ]| + |[ 0,0,(p1+p2).3 ]|;
 |[ (p1+p2).1,(p1+p2).2,0 ]|.1 = (p1+p2).1 &
     |[ (p1+p2).1,(p1+p2).2,0 ]|.2 = (p1+p2).2 &
     |[ (p1+p2).1,(p1+p2).2,0 ]|.3 = 0;
 then (p1+p2).1*<e1> + (p1+p2).2*<e2> + (p1+p2).3*<e3>
   = |[ (p1+p2).1+0,(p1+p2).2+0,0 +(p1+p2).3]| by A10,A11,Lm2
  .= |[ (p1+p2).1,(p1+p2).2,(p1+p2).3 ]|;
     hence thesis by A4,A5,A6,Th1;
end;
