Nuprl Lemma : is-dag-append
∀[T:Type]. ∀[g1,g2:LabeledGraph(T)]. (is-dag(lg-append(g1;g2))) supposing (is-dag(g2) and is-dag(g1))
Proof
Definitions occuring in Statement :
is-dag: is-dag(g),
lg-append: lg-append(g1;g2),
labeled-graph: LabeledGraph(T),
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
universe: Type
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
is-dag: is-dag(g),
all: ∀x:A. B[x],
implies: P ⇒ Q,
prop: ℙ,
subtype_rel: A ⊆r B,
int_seg: {i..j-},
nat: ℕ,
lelt: i ≤ j < k,
and: P ∧ Q,
guard: {T},
decidable: Dec(P),
or: P ∨ Q,
false: False,
uiff: uiff(P;Q),
satisfiable_int_formula: satisfiable_int_formula(fmla),
exists: ∃x:A. B[x],
not: ¬A,
top: Top,
le: A ≤ B,
less_than: a < b,
iff: P ⇐⇒ Q
Latex:
\mforall{}[T:Type]. \mforall{}[g1,g2:LabeledGraph(T)].
(is-dag(lg-append(g1;g2))) supposing (is-dag(g2) and is-dag(g1))
Date html generated:
2016_05_17-AM-10_11_23
Last ObjectModification:
2016_01_18-AM-00_22_27
Theory : process-model
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