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 "> )