Metamath Proof Explorer


Theorem pw2gt0divsd

Description: Division of a positive surreal by a power of two. (Contributed by Scott Fenton, 7-Nov-2025)

Ref Expression
Hypotheses pw2gt0divsd.1 ( 𝜑𝐴 No )
pw2gt0divsd.2 ( 𝜑𝑁 ∈ ℕ0s )
Assertion pw2gt0divsd ( 𝜑 → ( 0s <s 𝐴 ↔ 0s <s ( 𝐴 /su ( 2ss 𝑁 ) ) ) )

Proof

Step Hyp Ref Expression
1 pw2gt0divsd.1 ( 𝜑𝐴 No )
2 pw2gt0divsd.2 ( 𝜑𝑁 ∈ ℕ0s )
3 0sno 0s No
4 3 a1i ( 𝜑 → 0s No )
5 1 2 pw2divscld ( 𝜑 → ( 𝐴 /su ( 2ss 𝑁 ) ) ∈ No )
6 2sno 2s No
7 expscl ( ( 2s No 𝑁 ∈ ℕ0s ) → ( 2ss 𝑁 ) ∈ No )
8 6 2 7 sylancr ( 𝜑 → ( 2ss 𝑁 ) ∈ No )
9 2nns 2s ∈ ℕs
10 nnsgt0 ( 2s ∈ ℕs → 0s <s 2s )
11 9 10 ax-mp 0s <s 2s
12 expsgt0 ( ( 2s No 𝑁 ∈ ℕ0s ∧ 0s <s 2s ) → 0s <s ( 2ss 𝑁 ) )
13 6 11 12 mp3an13 ( 𝑁 ∈ ℕ0s → 0s <s ( 2ss 𝑁 ) )
14 2 13 syl ( 𝜑 → 0s <s ( 2ss 𝑁 ) )
15 4 5 8 14 sltmul2d ( 𝜑 → ( 0s <s ( 𝐴 /su ( 2ss 𝑁 ) ) ↔ ( ( 2ss 𝑁 ) ·s 0s ) <s ( ( 2ss 𝑁 ) ·s ( 𝐴 /su ( 2ss 𝑁 ) ) ) ) )
16 muls01 ( ( 2ss 𝑁 ) ∈ No → ( ( 2ss 𝑁 ) ·s 0s ) = 0s )
17 8 16 syl ( 𝜑 → ( ( 2ss 𝑁 ) ·s 0s ) = 0s )
18 1 2 pw2divscan2d ( 𝜑 → ( ( 2ss 𝑁 ) ·s ( 𝐴 /su ( 2ss 𝑁 ) ) ) = 𝐴 )
19 17 18 breq12d ( 𝜑 → ( ( ( 2ss 𝑁 ) ·s 0s ) <s ( ( 2ss 𝑁 ) ·s ( 𝐴 /su ( 2ss 𝑁 ) ) ) ↔ 0s <s 𝐴 ) )
20 15 19 bitr2d ( 𝜑 → ( 0s <s 𝐴 ↔ 0s <s ( 𝐴 /su ( 2ss 𝑁 ) ) ) )