Theorem decbin3

Description: Decompose base 4 into base 2. (Contributed by Mario Carneiro, 18-Feb-2014)

	Ref	Expression
Hypothesis	decbin.1	|- A e. NN0
Assertion	decbin3	|- ( ( 4 x. A ) + 3 ) = ( ( 2 x. ( ( 2 x. A ) + 1 ) ) + 1 )

Proof

Step	Hyp	Ref	Expression
1		decbin.1	|- A e. NN0
2		4nn0	|- 4 e. NN0
3		2nn0	|- 2 e. NN0
4		2p1e3	|- ( 2 + 1 ) = 3
5	1	decbin2	|- ( ( 4 x. A ) + 2 ) = ( 2 x. ( ( 2 x. A ) + 1 ) )
6	5	eqcomi	|- ( 2 x. ( ( 2 x. A ) + 1 ) ) = ( ( 4 x. A ) + 2 )
7	2 1 3 4 6	numsuc	|- ( ( 2 x. ( ( 2 x. A ) + 1 ) ) + 1 ) = ( ( 4 x. A ) + 3 )
8	7	eqcomi	|- ( ( 4 x. A ) + 3 ) = ( ( 2 x. ( ( 2 x. A ) + 1 ) ) + 1 )