Metamath Proof Explorer


Theorem pw2divsidd

Description: Identity law for division over powers of two. (Contributed by Scott Fenton, 21-Feb-2026)

Ref Expression
Hypothesis pw2divs0d.1
|- ( ph -> N e. NN0_s )
Assertion pw2divsidd
|- ( ph -> ( ( 2s ^su N ) /su ( 2s ^su N ) ) = 1s )

Proof

Step Hyp Ref Expression
1 pw2divs0d.1
 |-  ( ph -> N e. NN0_s )
2 2sno
 |-  2s e. No
3 expscl
 |-  ( ( 2s e. No /\ N e. NN0_s ) -> ( 2s ^su N ) e. No )
4 2 1 3 sylancr
 |-  ( ph -> ( 2s ^su N ) e. No )
5 4 mulsridd
 |-  ( ph -> ( ( 2s ^su N ) x.s 1s ) = ( 2s ^su N ) )
6 1sno
 |-  1s e. No
7 6 a1i
 |-  ( ph -> 1s e. No )
8 4 7 1 pw2divsmuld
 |-  ( ph -> ( ( ( 2s ^su N ) /su ( 2s ^su N ) ) = 1s <-> ( ( 2s ^su N ) x.s 1s ) = ( 2s ^su N ) ) )
9 5 8 mpbird
 |-  ( ph -> ( ( 2s ^su N ) /su ( 2s ^su N ) ) = 1s )