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 ˙