Metamath Proof Explorer


Theorem itgoss

Description: An integral element is integral over a subset. (Contributed by Stefan O'Rear, 27-Nov-2014)

Ref Expression
Assertion itgoss ( ( 𝑆𝑇𝑇 ⊆ ℂ ) → ( IntgOver ‘ 𝑆 ) ⊆ ( IntgOver ‘ 𝑇 ) )

Proof

Step Hyp Ref Expression
1 plyss ( ( 𝑆𝑇𝑇 ⊆ ℂ ) → ( Poly ‘ 𝑆 ) ⊆ ( Poly ‘ 𝑇 ) )
2 ssrexv ( ( Poly ‘ 𝑆 ) ⊆ ( Poly ‘ 𝑇 ) → ( ∃ 𝑏 ∈ ( Poly ‘ 𝑆 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) → ∃ 𝑏 ∈ ( Poly ‘ 𝑇 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) ) )
3 1 2 syl ( ( 𝑆𝑇𝑇 ⊆ ℂ ) → ( ∃ 𝑏 ∈ ( Poly ‘ 𝑆 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) → ∃ 𝑏 ∈ ( Poly ‘ 𝑇 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) ) )
4 3 ralrimivw ( ( 𝑆𝑇𝑇 ⊆ ℂ ) → ∀ 𝑎 ∈ ℂ ( ∃ 𝑏 ∈ ( Poly ‘ 𝑆 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) → ∃ 𝑏 ∈ ( Poly ‘ 𝑇 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) ) )
5 ss2rab ( { 𝑎 ∈ ℂ ∣ ∃ 𝑏 ∈ ( Poly ‘ 𝑆 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) } ⊆ { 𝑎 ∈ ℂ ∣ ∃ 𝑏 ∈ ( Poly ‘ 𝑇 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) } ↔ ∀ 𝑎 ∈ ℂ ( ∃ 𝑏 ∈ ( Poly ‘ 𝑆 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) → ∃ 𝑏 ∈ ( Poly ‘ 𝑇 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) ) )
6 4 5 sylibr ( ( 𝑆𝑇𝑇 ⊆ ℂ ) → { 𝑎 ∈ ℂ ∣ ∃ 𝑏 ∈ ( Poly ‘ 𝑆 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) } ⊆ { 𝑎 ∈ ℂ ∣ ∃ 𝑏 ∈ ( Poly ‘ 𝑇 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) } )
7 sstr ( ( 𝑆𝑇𝑇 ⊆ ℂ ) → 𝑆 ⊆ ℂ )
8 itgoval ( 𝑆 ⊆ ℂ → ( IntgOver ‘ 𝑆 ) = { 𝑎 ∈ ℂ ∣ ∃ 𝑏 ∈ ( Poly ‘ 𝑆 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) } )
9 7 8 syl ( ( 𝑆𝑇𝑇 ⊆ ℂ ) → ( IntgOver ‘ 𝑆 ) = { 𝑎 ∈ ℂ ∣ ∃ 𝑏 ∈ ( Poly ‘ 𝑆 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) } )
10 itgoval ( 𝑇 ⊆ ℂ → ( IntgOver ‘ 𝑇 ) = { 𝑎 ∈ ℂ ∣ ∃ 𝑏 ∈ ( Poly ‘ 𝑇 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) } )
11 10 adantl ( ( 𝑆𝑇𝑇 ⊆ ℂ ) → ( IntgOver ‘ 𝑇 ) = { 𝑎 ∈ ℂ ∣ ∃ 𝑏 ∈ ( Poly ‘ 𝑇 ) ( ( 𝑏𝑎 ) = 0 ∧ ( ( coeff ‘ 𝑏 ) ‘ ( deg ‘ 𝑏 ) ) = 1 ) } )
12 6 9 11 3sstr4d ( ( 𝑆𝑇𝑇 ⊆ ℂ ) → ( IntgOver ‘ 𝑆 ) ⊆ ( IntgOver ‘ 𝑇 ) )