Metamath Proof Explorer


Theorem decbin3

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

Ref Expression
Hypothesis decbin.1 𝐴 ∈ ℕ0
Assertion decbin3 ( ( 4 · 𝐴 ) + 3 ) = ( ( 2 · ( ( 2 · 𝐴 ) + 1 ) ) + 1 )

Proof

Step Hyp Ref Expression
1 decbin.1 𝐴 ∈ ℕ0
2 4nn0 4 ∈ ℕ0
3 2nn0 2 ∈ ℕ0
4 2p1e3 ( 2 + 1 ) = 3
5 1 decbin2 ( ( 4 · 𝐴 ) + 2 ) = ( 2 · ( ( 2 · 𝐴 ) + 1 ) )
6 5 eqcomi ( 2 · ( ( 2 · 𝐴 ) + 1 ) ) = ( ( 4 · 𝐴 ) + 2 )
7 2 1 3 4 6 numsuc ( ( 2 · ( ( 2 · 𝐴 ) + 1 ) ) + 1 ) = ( ( 4 · 𝐴 ) + 3 )
8 7 eqcomi ( ( 4 · 𝐴 ) + 3 ) = ( ( 2 · ( ( 2 · 𝐴 ) + 1 ) ) + 1 )