Metamath Proof Explorer


Theorem divsrecd

Description: Relationship between surreal division and reciprocal. (Contributed by Scott Fenton, 13-Aug-2025)

Ref Expression
Hypotheses divsrecd.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
divsrecd.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
divsrecd.3 ⊢ ( 𝜑 → 𝐵 ≠ 0s )
Assertion divsrecd ( 𝜑 → ( 𝐴 /su 𝐵 ) = ( 𝐴 ·s ( 1s /su 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 divsrecd.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
2 divsrecd.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
3 divsrecd.3 ⊢ ( 𝜑 → 𝐵 ≠ 0s )
4 1no ⊢ 1s ∈ No
5 4 a1i ⊢ ( 𝜑 → 1s ∈ No )
6 5 2 3 divscld ⊢ ( 𝜑 → ( 1s /su 𝐵 ) ∈ No )
7 2 1 6 muls12d ⊢ ( 𝜑 → ( 𝐵 ·s ( 𝐴 ·s ( 1s /su 𝐵 ) ) ) = ( 𝐴 ·s ( 𝐵 ·s ( 1s /su 𝐵 ) ) ) )
8 5 2 3 divscan2d ⊢ ( 𝜑 → ( 𝐵 ·s ( 1s /su 𝐵 ) ) = 1s )
9 8 oveq2d ⊢ ( 𝜑 → ( 𝐴 ·s ( 𝐵 ·s ( 1s /su 𝐵 ) ) ) = ( 𝐴 ·s 1s ) )
10 1 mulsridd ⊢ ( 𝜑 → ( 𝐴 ·s 1s ) = 𝐴 )
11 7 9 10 3eqtrd ⊢ ( 𝜑 → ( 𝐵 ·s ( 𝐴 ·s ( 1s /su 𝐵 ) ) ) = 𝐴 )
12 1 6 mulscld ⊢ ( 𝜑 → ( 𝐴 ·s ( 1s /su 𝐵 ) ) ∈ No )
13 1 12 2 3 divmulsd ⊢ ( 𝜑 → ( ( 𝐴 /su 𝐵 ) = ( 𝐴 ·s ( 1s /su 𝐵 ) ) ↔ ( 𝐵 ·s ( 𝐴 ·s ( 1s /su 𝐵 ) ) ) = 𝐴 ) )
14 11 13 mpbird ⊢ ( 𝜑 → ( 𝐴 /su 𝐵 ) = ( 𝐴 ·s ( 1s /su 𝐵 ) ) )