Metamath Proof Explorer


Theorem pw2divsassd

Description: An associative law for division by powers of two. (Contributed by Scott Fenton, 11-Dec-2025)

Ref Expression
Hypotheses pw2divsassd.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
pw2divsassd.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
pw2divsassd.3 ⊢ ( 𝜑 → 𝑁 ∈ ℕ0s )
Assertion pw2divsassd ( 𝜑 → ( ( 𝐴 ·s 𝐵 ) /su ( 2s ↑s 𝑁 ) ) = ( 𝐴 ·s ( 𝐵 /su ( 2s ↑s 𝑁 ) ) ) )

Proof

Step Hyp Ref Expression
1 pw2divsassd.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
2 pw2divsassd.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
3 pw2divsassd.3 ⊢ ( 𝜑 → 𝑁 ∈ ℕ0s )
4 2no ⊢ 2s ∈ No
5 expscl ⊢ ( ( 2s ∈ No ∧ 𝑁 ∈ ℕ0s ) → ( 2s ↑s 𝑁 ) ∈ No )
6 4 3 5 sylancr ⊢ ( 𝜑 → ( 2s ↑s 𝑁 ) ∈ No )
7 2ne0s ⊢ 2s ≠ 0s
8 expsne0 ⊢ ( ( 2s ∈ No ∧ 2s ≠ 0s ∧ 𝑁 ∈ ℕ0s ) → ( 2s ↑s 𝑁 ) ≠ 0s )
9 4 7 3 8 mp3an12i ⊢ ( 𝜑 → ( 2s ↑s 𝑁 ) ≠ 0s )
10 pw2recs ⊢ ( 𝑁 ∈ ℕ0s → ∃ 𝑥 ∈ No ( ( 2s ↑s 𝑁 ) ·s 𝑥 ) = 1s )
11 3 10 syl ⊢ ( 𝜑 → ∃ 𝑥 ∈ No ( ( 2s ↑s 𝑁 ) ·s 𝑥 ) = 1s )
12 1 2 6 9 11 divsasswd ⊢ ( 𝜑 → ( ( 𝐴 ·s 𝐵 ) /su ( 2s ↑s 𝑁 ) ) = ( 𝐴 ·s ( 𝐵 /su ( 2s ↑s 𝑁 ) ) ) )