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 } ) )