This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

Metamath Proof Explorer

Theorem s5eqd

Description: Equality theorem for a length 5 word. (Contributed by Mario Carneiro, 27-Feb-2016)

Ref Expression
Hypotheses s2eqd.1 
|- ( ph -> A = N )
s2eqd.2 
|- ( ph -> B = O )
s3eqd.3 
|- ( ph -> C = P )
s4eqd.4 
|- ( ph -> D = Q )
s5eqd.5 
|- ( ph -> E = R )
Assertion s5eqd 
|- ( ph -> <" A B C D E "> = <" N O P Q R "> )

Proof

Step Hyp Ref Expression
1 s2eqd.1 
 |-  ( ph -> A = N )
2 s2eqd.2 
 |-  ( ph -> B = O )
3 s3eqd.3 
 |-  ( ph -> C = P )
4 s4eqd.4 
 |-  ( ph -> D = Q )
5 s5eqd.5 
 |-  ( ph -> E = R )
6 1 2 3 4 s4eqd 
 |-  ( ph -> <" A B C D "> = <" N O P Q "> )
7 5 s1eqd 
 |-  ( ph -> <" E "> = <" R "> )
8 6 7 oveq12d 
 |-  ( ph -> ( <" A B C D "> ++ <" E "> ) = ( <" N O P Q "> ++ <" R "> ) )
9 df-s5 
 |-  <" A B C D E "> = ( <" A B C D "> ++ <" E "> )
10 df-s5 
 |-  <" N O P Q R "> = ( <" N O P Q "> ++ <" R "> )
11 8 9 10 3eqtr4g 
 |-  ( ph -> <" A B C D E "> = <" N O P Q R "> )