Metamath Proof Explorer


Theorem dibvalrel

Description: The value of partial isomorphism B is a relation. (Contributed by NM, 8-Mar-2014)

Ref Expression
Hypotheses dibcl.h ⊢ H = LHyp ⁡ K
dibcl.i ⊢ I = DIsoB ⁡ K ⁡ W
Assertion dibvalrel ⊢ K ∈ V ∧ W ∈ H → Rel ⁡ I ⁡ X

Proof

Step Hyp Ref Expression
1 dibcl.h ⊢ H = LHyp ⁡ K
2 dibcl.i ⊢ I = DIsoB ⁡ K ⁡ W
3 relxp ⊢ Rel ⁡ DIsoA ⁡ K ⁡ W ⁡ X × h ∈ LTrn ⁡ K ⁡ W ⟼ I ↾ Base K
4 eqid ⊢ DIsoA ⁡ K ⁡ W = DIsoA ⁡ K ⁡ W
5 1 4 2 dibdiadm ⊢ K ∈ V ∧ W ∈ H → dom ⁡ I = dom ⁡ DIsoA ⁡ K ⁡ W
6 5 eleq2d ⊢ K ∈ V ∧ W ∈ H → X ∈ dom ⁡ I ↔ X ∈ dom ⁡ DIsoA ⁡ K ⁡ W
7 6 biimpa ⊢ K ∈ V ∧ W ∈ H ∧ X ∈ dom ⁡ I → X ∈ dom ⁡ DIsoA ⁡ K ⁡ W
8 eqid ⊢ Base K = Base K
9 eqid ⊢ LTrn ⁡ K ⁡ W = LTrn ⁡ K ⁡ W
10 eqid ⊢ h ∈ LTrn ⁡ K ⁡ W ⟼ I ↾ Base K = h ∈ LTrn ⁡ K ⁡ W ⟼ I ↾ Base K
11 8 1 9 10 4 2 dibval ⊢ K ∈ V ∧ W ∈ H ∧ X ∈ dom ⁡ DIsoA ⁡ K ⁡ W → I ⁡ X = DIsoA ⁡ K ⁡ W ⁡ X × h ∈ LTrn ⁡ K ⁡ W ⟼ I ↾ Base K
12 7 11 syldan ⊢ K ∈ V ∧ W ∈ H ∧ X ∈ dom ⁡ I → I ⁡ X = DIsoA ⁡ K ⁡ W ⁡ X × h ∈ LTrn ⁡ K ⁡ W ⟼ I ↾ Base K
13 12 releqd ⊢ K ∈ V ∧ W ∈ H ∧ X ∈ dom ⁡ I → Rel ⁡ I ⁡ X ↔ Rel ⁡ DIsoA ⁡ K ⁡ W ⁡ X × h ∈ LTrn ⁡ K ⁡ W ⟼ I ↾ Base K
14 3 13 mpbiri ⊢ K ∈ V ∧ W ∈ H ∧ X ∈ dom ⁡ I → Rel ⁡ I ⁡ X
15 rel0 ⊢ Rel ⁡ ∅
16 ndmfv ⊢ ¬ X ∈ dom ⁡ I → I ⁡ X = ∅
17 16 releqd ⊢ ¬ X ∈ dom ⁡ I → Rel ⁡ I ⁡ X ↔ Rel ⁡ ∅
18 15 17 mpbiri ⊢ ¬ X ∈ dom ⁡ I → Rel ⁡ I ⁡ X
19 18 adantl ⊢ K ∈ V ∧ W ∈ H ∧ ¬ X ∈ dom ⁡ I → Rel ⁡ I ⁡ X
20 14 19 pm2.61dan ⊢ K ∈ V ∧ W ∈ H → Rel ⁡ I ⁡ X