Metamath Proof Explorer


Theorem divmulswd

Description: Relationship between surreal division and multiplication. Weak version that does not assume reciprocals. (Contributed by Scott Fenton, 12-Mar-2025)

Ref Expression
Hypotheses divmulswd.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
divmulswd.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
divmulswd.3 ⊢ ( 𝜑 → 𝐶 ∈ No )
divmulswd.4 ⊢ ( 𝜑 → 𝐶 ≠ 0s )
divmulswd.5 ⊢ ( 𝜑 → ∃ 𝑥 ∈ No ( 𝐶 ·s 𝑥 ) = 1s )
Assertion divmulswd ( 𝜑 → ( ( 𝐴 /su 𝐶 ) = 𝐵 ↔ ( 𝐶 ·s 𝐵 ) = 𝐴 ) )

Proof

Step Hyp Ref Expression
1 divmulswd.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
2 divmulswd.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
3 divmulswd.3 ⊢ ( 𝜑 → 𝐶 ∈ No )
4 divmulswd.4 ⊢ ( 𝜑 → 𝐶 ≠ 0s )
5 divmulswd.5 ⊢ ( 𝜑 → ∃ 𝑥 ∈ No ( 𝐶 ·s 𝑥 ) = 1s )
6 3 4 jca ⊢ ( 𝜑 → ( 𝐶 ∈ No ∧ 𝐶 ≠ 0s ) )
7 divmulsw ⊢ ( ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ ( 𝐶 ∈ No ∧ 𝐶 ≠ 0s ) ) ∧ ∃ 𝑥 ∈ No ( 𝐶 ·s 𝑥 ) = 1s ) → ( ( 𝐴 /su 𝐶 ) = 𝐵 ↔ ( 𝐶 ·s 𝐵 ) = 𝐴 ) )
8 1 2 6 5 7 syl31anc ⊢ ( 𝜑 → ( ( 𝐴 /su 𝐶 ) = 𝐵 ↔ ( 𝐶 ·s 𝐵 ) = 𝐴 ) )