Metamath Proof Explorer


Theorem hoidifhspval2

Description: D is a function that returns the representation of the left side of the difference of a half-open interval and a half-space. Used in Lemma 115F of Fremlin1 p. 31 . (Contributed by Glauco Siliprandi, 24-Dec-2020)

Ref Expression
Hypotheses hoidifhspval2.d 𝐷 = ( 𝑥 ∈ ℝ ↦ ( 𝑎 ∈ ( ℝ ↑m 𝑋 ) ↦ ( 𝑘𝑋 ↦ if ( 𝑘 = 𝐾 , if ( 𝑥 ≤ ( 𝑎𝑘 ) , ( 𝑎𝑘 ) , 𝑥 ) , ( 𝑎𝑘 ) ) ) ) )
hoidifhspval2.y ( 𝜑𝑌 ∈ ℝ )
hoidifhspval2.x ( 𝜑𝑋𝑉 )
hoidifhspval2.a ( 𝜑𝐴 : 𝑋 ⟶ ℝ )
Assertion hoidifhspval2 ( 𝜑 → ( ( 𝐷𝑌 ) ‘ 𝐴 ) = ( 𝑘𝑋 ↦ if ( 𝑘 = 𝐾 , if ( 𝑌 ≤ ( 𝐴𝑘 ) , ( 𝐴𝑘 ) , 𝑌 ) , ( 𝐴𝑘 ) ) ) )

Proof

Step Hyp Ref Expression
1 hoidifhspval2.d 𝐷 = ( 𝑥 ∈ ℝ ↦ ( 𝑎 ∈ ( ℝ ↑m 𝑋 ) ↦ ( 𝑘𝑋 ↦ if ( 𝑘 = 𝐾 , if ( 𝑥 ≤ ( 𝑎𝑘 ) , ( 𝑎𝑘 ) , 𝑥 ) , ( 𝑎𝑘 ) ) ) ) )
2 hoidifhspval2.y ( 𝜑𝑌 ∈ ℝ )
3 hoidifhspval2.x ( 𝜑𝑋𝑉 )
4 hoidifhspval2.a ( 𝜑𝐴 : 𝑋 ⟶ ℝ )
5 1 2 hoidifhspval ( 𝜑 → ( 𝐷𝑌 ) = ( 𝑎 ∈ ( ℝ ↑m 𝑋 ) ↦ ( 𝑘𝑋 ↦ if ( 𝑘 = 𝐾 , if ( 𝑌 ≤ ( 𝑎𝑘 ) , ( 𝑎𝑘 ) , 𝑌 ) , ( 𝑎𝑘 ) ) ) ) )
6 fveq1 ( 𝑎 = 𝐴 → ( 𝑎𝑘 ) = ( 𝐴𝑘 ) )
7 6 breq2d ( 𝑎 = 𝐴 → ( 𝑌 ≤ ( 𝑎𝑘 ) ↔ 𝑌 ≤ ( 𝐴𝑘 ) ) )
8 7 6 ifbieq1d ( 𝑎 = 𝐴 → if ( 𝑌 ≤ ( 𝑎𝑘 ) , ( 𝑎𝑘 ) , 𝑌 ) = if ( 𝑌 ≤ ( 𝐴𝑘 ) , ( 𝐴𝑘 ) , 𝑌 ) )
9 8 6 ifeq12d ( 𝑎 = 𝐴 → if ( 𝑘 = 𝐾 , if ( 𝑌 ≤ ( 𝑎𝑘 ) , ( 𝑎𝑘 ) , 𝑌 ) , ( 𝑎𝑘 ) ) = if ( 𝑘 = 𝐾 , if ( 𝑌 ≤ ( 𝐴𝑘 ) , ( 𝐴𝑘 ) , 𝑌 ) , ( 𝐴𝑘 ) ) )
10 9 mpteq2dv ( 𝑎 = 𝐴 → ( 𝑘𝑋 ↦ if ( 𝑘 = 𝐾 , if ( 𝑌 ≤ ( 𝑎𝑘 ) , ( 𝑎𝑘 ) , 𝑌 ) , ( 𝑎𝑘 ) ) ) = ( 𝑘𝑋 ↦ if ( 𝑘 = 𝐾 , if ( 𝑌 ≤ ( 𝐴𝑘 ) , ( 𝐴𝑘 ) , 𝑌 ) , ( 𝐴𝑘 ) ) ) )
11 10 adantl ( ( 𝜑𝑎 = 𝐴 ) → ( 𝑘𝑋 ↦ if ( 𝑘 = 𝐾 , if ( 𝑌 ≤ ( 𝑎𝑘 ) , ( 𝑎𝑘 ) , 𝑌 ) , ( 𝑎𝑘 ) ) ) = ( 𝑘𝑋 ↦ if ( 𝑘 = 𝐾 , if ( 𝑌 ≤ ( 𝐴𝑘 ) , ( 𝐴𝑘 ) , 𝑌 ) , ( 𝐴𝑘 ) ) ) )
12 reex ℝ ∈ V
13 12 a1i ( 𝜑 → ℝ ∈ V )
14 13 3 jca ( 𝜑 → ( ℝ ∈ V ∧ 𝑋𝑉 ) )
15 elmapg ( ( ℝ ∈ V ∧ 𝑋𝑉 ) → ( 𝐴 ∈ ( ℝ ↑m 𝑋 ) ↔ 𝐴 : 𝑋 ⟶ ℝ ) )
16 14 15 syl ( 𝜑 → ( 𝐴 ∈ ( ℝ ↑m 𝑋 ) ↔ 𝐴 : 𝑋 ⟶ ℝ ) )
17 4 16 mpbird ( 𝜑𝐴 ∈ ( ℝ ↑m 𝑋 ) )
18 mptexg ( 𝑋𝑉 → ( 𝑘𝑋 ↦ if ( 𝑘 = 𝐾 , if ( 𝑌 ≤ ( 𝐴𝑘 ) , ( 𝐴𝑘 ) , 𝑌 ) , ( 𝐴𝑘 ) ) ) ∈ V )
19 3 18 syl ( 𝜑 → ( 𝑘𝑋 ↦ if ( 𝑘 = 𝐾 , if ( 𝑌 ≤ ( 𝐴𝑘 ) , ( 𝐴𝑘 ) , 𝑌 ) , ( 𝐴𝑘 ) ) ) ∈ V )
20 5 11 17 19 fvmptd ( 𝜑 → ( ( 𝐷𝑌 ) ‘ 𝐴 ) = ( 𝑘𝑋 ↦ if ( 𝑘 = 𝐾 , if ( 𝑌 ≤ ( 𝐴𝑘 ) , ( 𝐴𝑘 ) , 𝑌 ) , ( 𝐴𝑘 ) ) ) )