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Metamath Proof Explorer

Theorem dibval

Description: The partial isomorphism B for a lattice K . (Contributed by NM, 8-Dec-2013)

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

Proof

Step Hyp Ref Expression
1 dibval.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 dibval.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
3 dibval.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
4 dibval.o ⊢ 0 = ( 𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵 ) )
5 dibval.j ⊢ 𝐽 = ( ( DIsoA ‘ 𝐾 ) ‘ 𝑊 )
6 dibval.i ⊢ 𝐼 = ( ( DIsoB ‘ 𝐾 ) ‘ 𝑊 )
7 1 2 3 4 5 6 dibfval ⊢ ( ( 𝐾 ∈ 𝑉 ∧ 𝑊 ∈ 𝐻 ) → 𝐼 = ( 𝑥 ∈ dom 𝐽 ↦ ( ( 𝐽 ‘ 𝑥 ) × { 0 } ) ) )
8 7 adantr ⊢ ( ( ( 𝐾 ∈ 𝑉 ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑋 ∈ dom 𝐽 ) → 𝐼 = ( 𝑥 ∈ dom 𝐽 ↦ ( ( 𝐽 ‘ 𝑥 ) × { 0 } ) ) )
9 8 fveq1d ⊢ ( ( ( 𝐾 ∈ 𝑉 ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑋 ∈ dom 𝐽 ) → ( 𝐼 ‘ 𝑋 ) = ( ( 𝑥 ∈ dom 𝐽 ↦ ( ( 𝐽 ‘ 𝑥 ) × { 0 } ) ) ‘ 𝑋 ) )
10 fveq2 ⊢ ( 𝑥 = 𝑋 → ( 𝐽 ‘ 𝑥 ) = ( 𝐽 ‘ 𝑋 ) )
11 10 xpeq1d ⊢ ( 𝑥 = 𝑋 → ( ( 𝐽 ‘ 𝑥 ) × { 0 } ) = ( ( 𝐽 ‘ 𝑋 ) × { 0 } ) )
12 eqid ⊢ ( 𝑥 ∈ dom 𝐽 ↦ ( ( 𝐽 ‘ 𝑥 ) × { 0 } ) ) = ( 𝑥 ∈ dom 𝐽 ↦ ( ( 𝐽 ‘ 𝑥 ) × { 0 } ) )
13 fvex ⊢ ( 𝐽 ‘ 𝑋 ) ∈ V
14 snex ⊢ { 0 } ∈ V
15 13 14 xpex ⊢ ( ( 𝐽 ‘ 𝑋 ) × { 0 } ) ∈ V
16 11 12 15 fvmpt ⊢ ( 𝑋 ∈ dom 𝐽 → ( ( 𝑥 ∈ dom 𝐽 ↦ ( ( 𝐽 ‘ 𝑥 ) × { 0 } ) ) ‘ 𝑋 ) = ( ( 𝐽 ‘ 𝑋 ) × { 0 } ) )
17 16 adantl ⊢ ( ( ( 𝐾 ∈ 𝑉 ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑋 ∈ dom 𝐽 ) → ( ( 𝑥 ∈ dom 𝐽 ↦ ( ( 𝐽 ‘ 𝑥 ) × { 0 } ) ) ‘ 𝑋 ) = ( ( 𝐽 ‘ 𝑋 ) × { 0 } ) )
18 9 17 eqtrd ⊢ ( ( ( 𝐾 ∈ 𝑉 ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑋 ∈ dom 𝐽 ) → ( 𝐼 ‘ 𝑋 ) = ( ( 𝐽 ‘ 𝑋 ) × { 0 } ) )