Metamath Proof Explorer


Theorem cutbdaybnd2

Description: An upper bound on the birthday of a surreal cut. (Contributed by Scott Fenton, 10-Dec-2021)

Ref Expression
Assertion cutbdaybnd2 ⊢ A ≪ s B → bday ⁡ A | s B ⊆ suc ⁡ ⋃ bday A ∪ B

Proof

Step Hyp Ref Expression
1 etaslts2 ⊢ A ≪ s B → ∃ x ∈ No A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B
2 cutbday ⊢ A ≪ s B → bday ⁡ A | s B = ⋂ bday y ∈ No | A ≪ s y ∧ y ≪ s B
3 2 adantr ⊢ A ≪ s B ∧ x ∈ No ∧ A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B → bday ⁡ A | s B = ⋂ bday y ∈ No | A ≪ s y ∧ y ≪ s B
4 bdayfn ⊢ bday Fn No
5 ssrab2 ⊢ y ∈ No | A ≪ s y ∧ y ≪ s B ⊆ No
6 simprl ⊢ A ≪ s B ∧ x ∈ No ∧ A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B → x ∈ No
7 simprr1 ⊢ A ≪ s B ∧ x ∈ No ∧ A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B → A ≪ s x
8 simprr2 ⊢ A ≪ s B ∧ x ∈ No ∧ A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B → x ≪ s B
9 7 8 jca ⊢ A ≪ s B ∧ x ∈ No ∧ A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B → A ≪ s x ∧ x ≪ s B
10 sneq ⊢ y = x → y = x
11 10 breq2d ⊢ y = x → A ≪ s y ↔ A ≪ s x
12 10 breq1d ⊢ y = x → y ≪ s B ↔ x ≪ s B
13 11 12 anbi12d ⊢ y = x → A ≪ s y ∧ y ≪ s B ↔ A ≪ s x ∧ x ≪ s B
14 13 elrab ⊢ x ∈ y ∈ No | A ≪ s y ∧ y ≪ s B ↔ x ∈ No ∧ A ≪ s x ∧ x ≪ s B
15 6 9 14 sylanbrc ⊢ A ≪ s B ∧ x ∈ No ∧ A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B → x ∈ y ∈ No | A ≪ s y ∧ y ≪ s B
16 fnfvima ⊢ bday Fn No ∧ y ∈ No | A ≪ s y ∧ y ≪ s B ⊆ No ∧ x ∈ y ∈ No | A ≪ s y ∧ y ≪ s B → bday ⁡ x ∈ bday y ∈ No | A ≪ s y ∧ y ≪ s B
17 4 5 15 16 mp3an12i ⊢ A ≪ s B ∧ x ∈ No ∧ A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B → bday ⁡ x ∈ bday y ∈ No | A ≪ s y ∧ y ≪ s B
18 intss1 ⊢ bday ⁡ x ∈ bday y ∈ No | A ≪ s y ∧ y ≪ s B → ⋂ bday y ∈ No | A ≪ s y ∧ y ≪ s B ⊆ bday ⁡ x
19 17 18 syl ⊢ A ≪ s B ∧ x ∈ No ∧ A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B → ⋂ bday y ∈ No | A ≪ s y ∧ y ≪ s B ⊆ bday ⁡ x
20 3 19 eqsstrd ⊢ A ≪ s B ∧ x ∈ No ∧ A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B → bday ⁡ A | s B ⊆ bday ⁡ x
21 simprr3 ⊢ A ≪ s B ∧ x ∈ No ∧ A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B → bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B
22 20 21 sstrd ⊢ A ≪ s B ∧ x ∈ No ∧ A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B → bday ⁡ A | s B ⊆ suc ⁡ ⋃ bday A ∪ B
23 22 rexlimdvaa ⊢ A ≪ s B → ∃ x ∈ No A ≪ s x ∧ x ≪ s B ∧ bday ⁡ x ⊆ suc ⁡ ⋃ bday A ∪ B → bday ⁡ A | s B ⊆ suc ⁡ ⋃ bday A ∪ B
24 1 23 mpd ⊢ A ≪ s B → bday ⁡ A | s B ⊆ suc ⁡ ⋃ bday A ∪ B