Metamath Proof Explorer


Theorem muls01

Description: Surreal multiplication by zero. (Contributed by Scott Fenton, 4-Feb-2025)

Ref Expression
Assertion muls01 ⊢ A ∈ No → A ⋅ s 0 s = 0 s

Proof

Step Hyp Ref Expression
1 0no ⊢ 0 s ∈ No
2 mulsval ⊢ A ∈ No ∧ 0 s ∈ No → A ⋅ s 0 s = a | ∃ p ∈ L ⁡ A ∃ q ∈ L ⁡ 0 s a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q ∪ b | ∃ r ∈ R ⁡ A ∃ s ∈ R ⁡ 0 s b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s | s c | ∃ t ∈ L ⁡ A ∃ u ∈ R ⁡ 0 s c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u ∪ d | ∃ v ∈ R ⁡ A ∃ w ∈ L ⁡ 0 s d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w
3 1 2 mpan2 ⊢ A ∈ No → A ⋅ s 0 s = a | ∃ p ∈ L ⁡ A ∃ q ∈ L ⁡ 0 s a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q ∪ b | ∃ r ∈ R ⁡ A ∃ s ∈ R ⁡ 0 s b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s | s c | ∃ t ∈ L ⁡ A ∃ u ∈ R ⁡ 0 s c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u ∪ d | ∃ v ∈ R ⁡ A ∃ w ∈ L ⁡ 0 s d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w
4 rex0 ⊢ ¬ ∃ q ∈ ∅ a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q
5 left0s ⊢ L ⁡ 0 s = ∅
6 5 rexeqi ⊢ ∃ q ∈ L ⁡ 0 s a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q ↔ ∃ q ∈ ∅ a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q
7 4 6 mtbir ⊢ ¬ ∃ q ∈ L ⁡ 0 s a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q
8 7 a1i ⊢ p ∈ L ⁡ A → ¬ ∃ q ∈ L ⁡ 0 s a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q
9 8 nrex ⊢ ¬ ∃ p ∈ L ⁡ A ∃ q ∈ L ⁡ 0 s a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q
10 9 abf ⊢ a | ∃ p ∈ L ⁡ A ∃ q ∈ L ⁡ 0 s a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q = ∅
11 rex0 ⊢ ¬ ∃ s ∈ ∅ b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s
12 right0s ⊢ R ⁡ 0 s = ∅
13 12 rexeqi ⊢ ∃ s ∈ R ⁡ 0 s b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s ↔ ∃ s ∈ ∅ b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s
14 11 13 mtbir ⊢ ¬ ∃ s ∈ R ⁡ 0 s b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s
15 14 a1i ⊢ r ∈ R ⁡ A → ¬ ∃ s ∈ R ⁡ 0 s b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s
16 15 nrex ⊢ ¬ ∃ r ∈ R ⁡ A ∃ s ∈ R ⁡ 0 s b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s
17 16 abf ⊢ b | ∃ r ∈ R ⁡ A ∃ s ∈ R ⁡ 0 s b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s = ∅
18 10 17 uneq12i ⊢ a | ∃ p ∈ L ⁡ A ∃ q ∈ L ⁡ 0 s a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q ∪ b | ∃ r ∈ R ⁡ A ∃ s ∈ R ⁡ 0 s b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s = ∅ ∪ ∅
19 un0 ⊢ ∅ ∪ ∅ = ∅
20 18 19 eqtri ⊢ a | ∃ p ∈ L ⁡ A ∃ q ∈ L ⁡ 0 s a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q ∪ b | ∃ r ∈ R ⁡ A ∃ s ∈ R ⁡ 0 s b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s = ∅
21 rex0 ⊢ ¬ ∃ u ∈ ∅ c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u
22 12 rexeqi ⊢ ∃ u ∈ R ⁡ 0 s c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u ↔ ∃ u ∈ ∅ c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u
23 21 22 mtbir ⊢ ¬ ∃ u ∈ R ⁡ 0 s c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u
24 23 a1i ⊢ t ∈ L ⁡ A → ¬ ∃ u ∈ R ⁡ 0 s c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u
25 24 nrex ⊢ ¬ ∃ t ∈ L ⁡ A ∃ u ∈ R ⁡ 0 s c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u
26 25 abf ⊢ c | ∃ t ∈ L ⁡ A ∃ u ∈ R ⁡ 0 s c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u = ∅
27 rex0 ⊢ ¬ ∃ w ∈ ∅ d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w
28 5 rexeqi ⊢ ∃ w ∈ L ⁡ 0 s d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w ↔ ∃ w ∈ ∅ d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w
29 27 28 mtbir ⊢ ¬ ∃ w ∈ L ⁡ 0 s d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w
30 29 a1i ⊢ v ∈ R ⁡ A → ¬ ∃ w ∈ L ⁡ 0 s d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w
31 30 nrex ⊢ ¬ ∃ v ∈ R ⁡ A ∃ w ∈ L ⁡ 0 s d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w
32 31 abf ⊢ d | ∃ v ∈ R ⁡ A ∃ w ∈ L ⁡ 0 s d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w = ∅
33 26 32 uneq12i ⊢ c | ∃ t ∈ L ⁡ A ∃ u ∈ R ⁡ 0 s c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u ∪ d | ∃ v ∈ R ⁡ A ∃ w ∈ L ⁡ 0 s d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w = ∅ ∪ ∅
34 33 19 eqtri ⊢ c | ∃ t ∈ L ⁡ A ∃ u ∈ R ⁡ 0 s c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u ∪ d | ∃ v ∈ R ⁡ A ∃ w ∈ L ⁡ 0 s d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w = ∅
35 20 34 oveq12i ⊢ a | ∃ p ∈ L ⁡ A ∃ q ∈ L ⁡ 0 s a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q ∪ b | ∃ r ∈ R ⁡ A ∃ s ∈ R ⁡ 0 s b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s | s c | ∃ t ∈ L ⁡ A ∃ u ∈ R ⁡ 0 s c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u ∪ d | ∃ v ∈ R ⁡ A ∃ w ∈ L ⁡ 0 s d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w = ∅ | s ∅
36 df-0s ⊢ 0 s = ∅ | s ∅
37 35 36 eqtr4i ⊢ a | ∃ p ∈ L ⁡ A ∃ q ∈ L ⁡ 0 s a = p ⋅ s 0 s + s A ⋅ s q - s p ⋅ s q ∪ b | ∃ r ∈ R ⁡ A ∃ s ∈ R ⁡ 0 s b = r ⋅ s 0 s + s A ⋅ s s - s r ⋅ s s | s c | ∃ t ∈ L ⁡ A ∃ u ∈ R ⁡ 0 s c = t ⋅ s 0 s + s A ⋅ s u - s t ⋅ s u ∪ d | ∃ v ∈ R ⁡ A ∃ w ∈ L ⁡ 0 s d = v ⋅ s 0 s + s A ⋅ s w - s v ⋅ s w = 0 s
38 3 37 eqtrdi ⊢ A ∈ No → A ⋅ s 0 s = 0 s