Metamath Proof Explorer


Theorem pw2divs0d

Description: Division into zero is zero for a power of two. (Contributed by Scott Fenton, 21-Feb-2026)

Ref Expression
Hypothesis pw2divs0d.1 ( 𝜑𝑁 ∈ ℕ0s )
Assertion pw2divs0d ( 𝜑 → ( 0s /su ( 2ss 𝑁 ) ) = 0s )

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 muls01 ( ( 2ss 𝑁 ) ∈ No → ( ( 2ss 𝑁 ) ·s 0s ) = 0s )
6 4 5 syl ( 𝜑 → ( ( 2ss 𝑁 ) ·s 0s ) = 0s )
7 0sno 0s No
8 7 a1i ( 𝜑 → 0s No )
9 8 8 1 pw2divsmuld ( 𝜑 → ( ( 0s /su ( 2ss 𝑁 ) ) = 0s ↔ ( ( 2ss 𝑁 ) ·s 0s ) = 0s ) )
10 6 9 mpbird ( 𝜑 → ( 0s /su ( 2ss 𝑁 ) ) = 0s )