Metamath Proof Explorer


Theorem cutminmax

Description: If the left set of X has a maximum and the right set of X has a minimum, then X is equal to the cut of the maximum and the minimum. (Contributed by Scott Fenton, 23-Feb-2026)

Ref Expression
Hypotheses cutminmax.1 ( 𝜑𝐿 ∈ ( L ‘ 𝑋 ) )
cutminmax.2 ( 𝜑 → ∀ 𝑥 ∈ ( L ‘ 𝑋 ) 𝑥 ≤s 𝐿 )
cutminmax.3 ( 𝜑𝑅 ∈ ( R ‘ 𝑋 ) )
cutminmax.4 ( 𝜑 → ∀ 𝑦 ∈ ( R ‘ 𝑋 ) 𝑅 ≤s 𝑦 )
Assertion cutminmax ( 𝜑𝑋 = ( { 𝐿 } |s { 𝑅 } ) )

Proof

Step Hyp Ref Expression
1 cutminmax.1 ( 𝜑𝐿 ∈ ( L ‘ 𝑋 ) )
2 cutminmax.2 ( 𝜑 → ∀ 𝑥 ∈ ( L ‘ 𝑋 ) 𝑥 ≤s 𝐿 )
3 cutminmax.3 ( 𝜑𝑅 ∈ ( R ‘ 𝑋 ) )
4 cutminmax.4 ( 𝜑 → ∀ 𝑦 ∈ ( R ‘ 𝑋 ) 𝑅 ≤s 𝑦 )
5 lltropt ( L ‘ 𝑋 ) <<s ( R ‘ 𝑋 )
6 5 a1i ( 𝜑 → ( L ‘ 𝑋 ) <<s ( R ‘ 𝑋 ) )
7 breq2 ( 𝑦 = 𝑏 → ( 𝑅 ≤s 𝑦𝑅 ≤s 𝑏 ) )
8 7 cbvralvw ( ∀ 𝑦 ∈ ( R ‘ 𝑋 ) 𝑅 ≤s 𝑦 ↔ ∀ 𝑏 ∈ ( R ‘ 𝑋 ) 𝑅 ≤s 𝑏 )
9 4 8 sylib ( 𝜑 → ∀ 𝑏 ∈ ( R ‘ 𝑋 ) 𝑅 ≤s 𝑏 )
10 6 3 9 cutmin ( 𝜑 → ( ( L ‘ 𝑋 ) |s ( R ‘ 𝑋 ) ) = ( ( L ‘ 𝑋 ) |s { 𝑅 } ) )
11 elfvdm ( 𝐿 ∈ ( L ‘ 𝑋 ) → 𝑋 ∈ dom L )
12 1 11 syl ( 𝜑𝑋 ∈ dom L )
13 leftf L : No ⟶ 𝒫 No
14 13 fdmi dom L = No
15 12 14 eleqtrdi ( 𝜑𝑋 No )
16 lrcut ( 𝑋 No → ( ( L ‘ 𝑋 ) |s ( R ‘ 𝑋 ) ) = 𝑋 )
17 15 16 syl ( 𝜑 → ( ( L ‘ 𝑋 ) |s ( R ‘ 𝑋 ) ) = 𝑋 )
18 3 snssd ( 𝜑 → { 𝑅 } ⊆ ( R ‘ 𝑋 ) )
19 sssslt2 ( ( ( L ‘ 𝑋 ) <<s ( R ‘ 𝑋 ) ∧ { 𝑅 } ⊆ ( R ‘ 𝑋 ) ) → ( L ‘ 𝑋 ) <<s { 𝑅 } )
20 5 18 19 sylancr ( 𝜑 → ( L ‘ 𝑋 ) <<s { 𝑅 } )
21 breq1 ( 𝑥 = 𝑎 → ( 𝑥 ≤s 𝐿𝑎 ≤s 𝐿 ) )
22 21 cbvralvw ( ∀ 𝑥 ∈ ( L ‘ 𝑋 ) 𝑥 ≤s 𝐿 ↔ ∀ 𝑎 ∈ ( L ‘ 𝑋 ) 𝑎 ≤s 𝐿 )
23 2 22 sylib ( 𝜑 → ∀ 𝑎 ∈ ( L ‘ 𝑋 ) 𝑎 ≤s 𝐿 )
24 20 1 23 cutmax ( 𝜑 → ( ( L ‘ 𝑋 ) |s { 𝑅 } ) = ( { 𝐿 } |s { 𝑅 } ) )
25 10 17 24 3eqtr3d ( 𝜑𝑋 = ( { 𝐿 } |s { 𝑅 } ) )