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Metamath Proof Explorer

Theorem dibval2

Description: Value of the partial isomorphism B. (Contributed by NM, 18-Jan-2014)

Ref Expression
Hypotheses dibval2.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
dibval2.l ⊢ ≤ = ( le ‘ 𝐾 )
dibval2.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
dibval2.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
dibval2.o ⊢ 0 = ( 𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵 ) )
dibval2.j ⊢ 𝐽 = ( ( DIsoA ‘ 𝐾 ) ‘ 𝑊 )
dibval2.i ⊢ 𝐼 = ( ( DIsoB ‘ 𝐾 ) ‘ 𝑊 )
Assertion dibval2 ( ( ( 𝐾 ∈ 𝑉 ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑋 ≤ 𝑊 ) ) → ( 𝐼 ‘ 𝑋 ) = ( ( 𝐽 ‘ 𝑋 ) × { 0 } ) )

Proof

Step Hyp Ref Expression
1 dibval2.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 dibval2.l ⊢ ≤ = ( le ‘ 𝐾 )
3 dibval2.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
4 dibval2.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
5 dibval2.o ⊢ 0 = ( 𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵 ) )
6 dibval2.j ⊢ 𝐽 = ( ( DIsoA ‘ 𝐾 ) ‘ 𝑊 )
7 dibval2.i ⊢ 𝐼 = ( ( DIsoB ‘ 𝐾 ) ‘ 𝑊 )
8 1 2 3 6 diaeldm ⊢ ( ( 𝐾 ∈ 𝑉 ∧ 𝑊 ∈ 𝐻 ) → ( 𝑋 ∈ dom 𝐽 ↔ ( 𝑋 ∈ 𝐵 ∧ 𝑋 ≤ 𝑊 ) ) )
9 8 biimpar ⊢ ( ( ( 𝐾 ∈ 𝑉 ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑋 ≤ 𝑊 ) ) → 𝑋 ∈ dom 𝐽 )
10 1 3 4 5 6 7 dibval ⊢ ( ( ( 𝐾 ∈ 𝑉 ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑋 ∈ dom 𝐽 ) → ( 𝐼 ‘ 𝑋 ) = ( ( 𝐽 ‘ 𝑋 ) × { 0 } ) )
11 9 10 syldan ⊢ ( ( ( 𝐾 ∈ 𝑉 ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑋 ≤ 𝑊 ) ) → ( 𝐼 ‘ 𝑋 ) = ( ( 𝐽 ‘ 𝑋 ) × { 0 } ) )