Metamath Proof Explorer


Theorem negsproplem1

Description: Lemma for surreal negation. We prove a pair of properties of surreal negation simultaneously. First, we instantiate some quantifiers. (Contributed by Scott Fenton, 2-Feb-2025)

Ref Expression
Hypotheses negsproplem.1 ⊢ φ → ∀ x ∈ No ∀ y ∈ No bday ⁡ x ∪ bday ⁡ y ∈ bday ⁡ A ∪ bday ⁡ B → + s ⁡ x ∈ No ∧ x < s y → + s ⁡ y < s + s ⁡ x
negsproplem1.1 ⊢ φ → X ∈ No
negsproplem1.2 ⊢ φ → Y ∈ No
negsproplem1.3 ⊢ φ → bday ⁡ X ∪ bday ⁡ Y ∈ bday ⁡ A ∪ bday ⁡ B
Assertion negsproplem1 ⊢ φ → + s ⁡ X ∈ No ∧ X < s Y → + s ⁡ Y < s + s ⁡ X

Proof

Step Hyp Ref Expression
1 negsproplem.1 ⊢ φ → ∀ x ∈ No ∀ y ∈ No bday ⁡ x ∪ bday ⁡ y ∈ bday ⁡ A ∪ bday ⁡ B → + s ⁡ x ∈ No ∧ x < s y → + s ⁡ y < s + s ⁡ x
2 negsproplem1.1 ⊢ φ → X ∈ No
3 negsproplem1.2 ⊢ φ → Y ∈ No
4 negsproplem1.3 ⊢ φ → bday ⁡ X ∪ bday ⁡ Y ∈ bday ⁡ A ∪ bday ⁡ B
5 2 3 jca ⊢ φ → X ∈ No ∧ Y ∈ No
6 fveq2 ⊢ x = X → bday ⁡ x = bday ⁡ X
7 6 uneq1d ⊢ x = X → bday ⁡ x ∪ bday ⁡ y = bday ⁡ X ∪ bday ⁡ y
8 7 eleq1d ⊢ x = X → bday ⁡ x ∪ bday ⁡ y ∈ bday ⁡ A ∪ bday ⁡ B ↔ bday ⁡ X ∪ bday ⁡ y ∈ bday ⁡ A ∪ bday ⁡ B
9 fveq2 ⊢ x = X → + s ⁡ x = + s ⁡ X
10 9 eleq1d ⊢ x = X → + s ⁡ x ∈ No ↔ + s ⁡ X ∈ No
11 breq1 ⊢ x = X → x < s y ↔ X < s y
12 9 breq2d ⊢ x = X → + s ⁡ y < s + s ⁡ x ↔ + s ⁡ y < s + s ⁡ X
13 11 12 imbi12d ⊢ x = X → x < s y → + s ⁡ y < s + s ⁡ x ↔ X < s y → + s ⁡ y < s + s ⁡ X
14 10 13 anbi12d ⊢ x = X → + s ⁡ x ∈ No ∧ x < s y → + s ⁡ y < s + s ⁡ x ↔ + s ⁡ X ∈ No ∧ X < s y → + s ⁡ y < s + s ⁡ X
15 8 14 imbi12d ⊢ x = X → bday ⁡ x ∪ bday ⁡ y ∈ bday ⁡ A ∪ bday ⁡ B → + s ⁡ x ∈ No ∧ x < s y → + s ⁡ y < s + s ⁡ x ↔ bday ⁡ X ∪ bday ⁡ y ∈ bday ⁡ A ∪ bday ⁡ B → + s ⁡ X ∈ No ∧ X < s y → + s ⁡ y < s + s ⁡ X
16 fveq2 ⊢ y = Y → bday ⁡ y = bday ⁡ Y
17 16 uneq2d ⊢ y = Y → bday ⁡ X ∪ bday ⁡ y = bday ⁡ X ∪ bday ⁡ Y
18 17 eleq1d ⊢ y = Y → bday ⁡ X ∪ bday ⁡ y ∈ bday ⁡ A ∪ bday ⁡ B ↔ bday ⁡ X ∪ bday ⁡ Y ∈ bday ⁡ A ∪ bday ⁡ B
19 breq2 ⊢ y = Y → X < s y ↔ X < s Y
20 fveq2 ⊢ y = Y → + s ⁡ y = + s ⁡ Y
21 20 breq1d ⊢ y = Y → + s ⁡ y < s + s ⁡ X ↔ + s ⁡ Y < s + s ⁡ X
22 19 21 imbi12d ⊢ y = Y → X < s y → + s ⁡ y < s + s ⁡ X ↔ X < s Y → + s ⁡ Y < s + s ⁡ X
23 22 anbi2d ⊢ y = Y → + s ⁡ X ∈ No ∧ X < s y → + s ⁡ y < s + s ⁡ X ↔ + s ⁡ X ∈ No ∧ X < s Y → + s ⁡ Y < s + s ⁡ X
24 18 23 imbi12d ⊢ y = Y → bday ⁡ X ∪ bday ⁡ y ∈ bday ⁡ A ∪ bday ⁡ B → + s ⁡ X ∈ No ∧ X < s y → + s ⁡ y < s + s ⁡ X ↔ bday ⁡ X ∪ bday ⁡ Y ∈ bday ⁡ A ∪ bday ⁡ B → + s ⁡ X ∈ No ∧ X < s Y → + s ⁡ Y < s + s ⁡ X
25 15 24 rspc2v ⊢ X ∈ No ∧ Y ∈ No → ∀ x ∈ No ∀ y ∈ No bday ⁡ x ∪ bday ⁡ y ∈ bday ⁡ A ∪ bday ⁡ B → + s ⁡ x ∈ No ∧ x < s y → + s ⁡ y < s + s ⁡ x → bday ⁡ X ∪ bday ⁡ Y ∈ bday ⁡ A ∪ bday ⁡ B → + s ⁡ X ∈ No ∧ X < s Y → + s ⁡ Y < s + s ⁡ X
26 5 1 4 25 syl3c ⊢ φ → + s ⁡ X ∈ No ∧ X < s Y → + s ⁡ Y < s + s ⁡ X