Metamath Proof Explorer


Theorem recsne0

Description: If a surreal has a reciprocal, then it is nonzero. (Contributed by Scott Fenton, 5-Sep-2025)

Ref Expression
Hypotheses recsne0.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
recsne0.2 ⊢ ( 𝜑 → ∃ 𝑥 ∈ No ( 𝐴 ·s 𝑥 ) = 1s )
Assertion recsne0 ( 𝜑 → 𝐴 ≠ 0s )

Proof

Step Hyp Ref Expression
1 recsne0.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
2 recsne0.2 ⊢ ( 𝜑 → ∃ 𝑥 ∈ No ( 𝐴 ·s 𝑥 ) = 1s )
3 oveq2 ⊢ ( 𝑥 = 𝑦 → ( 𝐴 ·s 𝑥 ) = ( 𝐴 ·s 𝑦 ) )
4 3 eqeq1d ⊢ ( 𝑥 = 𝑦 → ( ( 𝐴 ·s 𝑥 ) = 1s ↔ ( 𝐴 ·s 𝑦 ) = 1s ) )
5 4 cbvrexvw ⊢ ( ∃ 𝑥 ∈ No ( 𝐴 ·s 𝑥 ) = 1s ↔ ∃ 𝑦 ∈ No ( 𝐴 ·s 𝑦 ) = 1s )
6 2 5 sylib ⊢ ( 𝜑 → ∃ 𝑦 ∈ No ( 𝐴 ·s 𝑦 ) = 1s )
7 simprr ⊢ ( ( 𝜑 ∧ ( 𝑦 ∈ No ∧ ( 𝐴 ·s 𝑦 ) = 1s ) ) → ( 𝐴 ·s 𝑦 ) = 1s )
8 1ne0s ⊢ 1s ≠ 0s
9 8 a1i ⊢ ( ( 𝜑 ∧ ( 𝑦 ∈ No ∧ ( 𝐴 ·s 𝑦 ) = 1s ) ) → 1s ≠ 0s )
10 7 9 eqnetrd ⊢ ( ( 𝜑 ∧ ( 𝑦 ∈ No ∧ ( 𝐴 ·s 𝑦 ) = 1s ) ) → ( 𝐴 ·s 𝑦 ) ≠ 0s )
11 1 adantr ⊢ ( ( 𝜑 ∧ ( 𝑦 ∈ No ∧ ( 𝐴 ·s 𝑦 ) = 1s ) ) → 𝐴 ∈ No )
12 simprl ⊢ ( ( 𝜑 ∧ ( 𝑦 ∈ No ∧ ( 𝐴 ·s 𝑦 ) = 1s ) ) → 𝑦 ∈ No )
13 11 12 mulsne0bd ⊢ ( ( 𝜑 ∧ ( 𝑦 ∈ No ∧ ( 𝐴 ·s 𝑦 ) = 1s ) ) → ( ( 𝐴 ·s 𝑦 ) ≠ 0s ↔ ( 𝐴 ≠ 0s ∧ 𝑦 ≠ 0s ) ) )
14 10 13 mpbid ⊢ ( ( 𝜑 ∧ ( 𝑦 ∈ No ∧ ( 𝐴 ·s 𝑦 ) = 1s ) ) → ( 𝐴 ≠ 0s ∧ 𝑦 ≠ 0s ) )
15 14 simpld ⊢ ( ( 𝜑 ∧ ( 𝑦 ∈ No ∧ ( 𝐴 ·s 𝑦 ) = 1s ) ) → 𝐴 ≠ 0s )
16 6 15 rexlimddv ⊢ ( 𝜑 → 𝐴 ≠ 0s )