Step * 1 of Lemma free-dma-hom-is-lattice-hom

.....subterm..... T:t
1:n
1. Type
2. eq EqDecider(T)
3. dm BoundedDistributiveLattice
⊢ free-DeMorgan-lattice(T;eq) free-DeMorgan-algebra(T;eq) ∈ BoundedLatticeStructure
BY
(Assert free-DeMorgan-lattice(T;eq) ∈ BoundedLatticeStructure BY
         Auto) }

1
1. Type
2. eq EqDecider(T)
3. dm BoundedDistributiveLattice
4. free-DeMorgan-lattice(T;eq) ∈ BoundedLatticeStructure
⊢ free-DeMorgan-lattice(T;eq) free-DeMorgan-algebra(T;eq) ∈ BoundedLatticeStructure


Latex:


Latex:
.....subterm.....  T:t
1:n
1.  T  :  Type
2.  eq  :  EqDecider(T)
3.  dm  :  BoundedDistributiveLattice
\mvdash{}  free-DeMorgan-lattice(T;eq)  =  free-DeMorgan-algebra(T;eq)


By


Latex:
(Assert  free-DeMorgan-lattice(T;eq)  \mmember{}  BoundedLatticeStructure  BY
              Auto)




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