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 = 1 s
Assertion recsne0 ⊢ φ → A ≠ 0 s

Proof

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