Nuprl Lemma : is-dag-map
∀[T,S:Type]. ∀[f:T ⟶ S]. ∀[g:LabeledGraph(T)]. is-dag(lg-map(f;g)) supposing is-dag(g)
Proof
Definitions occuring in Statement :
lg-map: lg-map(f;g),
is-dag: is-dag(g),
labeled-graph: LabeledGraph(T),
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
function: 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],
int_seg: {i..j-},
lelt: i ≤ j < k,
and: P ∧ Q,
decidable: Dec(P),
or: P ∨ Q,
satisfiable_int_formula: satisfiable_int_formula(fmla),
exists: ∃x:A. B[x],
false: False,
implies: P ⇒ Q,
not: ¬A,
top: Top,
prop: ℙ,
le: A ≤ B,
less_than: a < b,
guard: {T},
iff: P ⇐⇒ Q,
subtype_rel: A ⊆r B,
nat: ℕ
Latex:
\mforall{}[T,S:Type]. \mforall{}[f:T {}\mrightarrow{} S]. \mforall{}[g:LabeledGraph(T)]. is-dag(lg-map(f;g)) supposing is-dag(g)
Date html generated:
2016_05_17-AM-10_12_35
Last ObjectModification:
2016_01_18-AM-00_21_47
Theory : process-model
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