Metamath Proof Explorer


Theorem s4eqd

Description: Equality theorem for a length 4 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 )
Assertion s4eqd
|- ( ph -> <" A B C D "> = <" N O P Q "> )

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 1 2 3 s3eqd
 |-  ( ph -> <" A B C "> = <" N O P "> )
6 4 s1eqd
 |-  ( ph -> <" D "> = <" Q "> )
7 5 6 oveq12d
 |-  ( ph -> ( <" A B C "> ++ <" D "> ) = ( <" N O P "> ++ <" Q "> ) )
8 df-s4
 |-  <" A B C D "> = ( <" A B C "> ++ <" D "> )
9 df-s4
 |-  <" N O P Q "> = ( <" N O P "> ++ <" Q "> )
10 7 8 9 3eqtr4g
 |-  ( ph -> <" A B C D "> = <" N O P Q "> )