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 ( 𝜑𝑁 ∈ ℕ0s )
Assertion pw2divsidd ( 𝜑 → ( ( 2ss 𝑁 ) /su ( 2ss 𝑁 ) ) = 1s )

Proof

Step Hyp Ref Expression
1 pw2divs0d.1 ( 𝜑𝑁 ∈ ℕ0s )
2 2sno 2s No
3 expscl ( ( 2s No 𝑁 ∈ ℕ0s ) → ( 2ss 𝑁 ) ∈ No )
4 2 1 3 sylancr ( 𝜑 → ( 2ss 𝑁 ) ∈ No )
5 4 mulsridd ( 𝜑 → ( ( 2ss 𝑁 ) ·s 1s ) = ( 2ss 𝑁 ) )
6 1sno 1s No
7 6 a1i ( 𝜑 → 1s No )
8 4 7 1 pw2divsmuld ( 𝜑 → ( ( ( 2ss 𝑁 ) /su ( 2ss 𝑁 ) ) = 1s ↔ ( ( 2ss 𝑁 ) ·s 1s ) = ( 2ss 𝑁 ) ) )
9 5 8 mpbird ( 𝜑 → ( ( 2ss 𝑁 ) /su ( 2ss 𝑁 ) ) = 1s )