Nuprl Lemma : concat-lifting-loc-3_wf
∀[A,B,C,D:Type]. ∀[f:Id ⟶ A ⟶ B ⟶ C ⟶ bag(D)].
(concat-lifting-loc-3(f) ∈ Id ⟶ bag(A) ⟶ bag(B) ⟶ bag(C) ⟶ bag(D))
Proof
Definitions occuring in Statement :
concat-lifting-loc-3: concat-lifting-loc-3(f),
Id: Id,
uall: ∀[x:A]. B[x],
member: t ∈ T,
function: x:A ⟶ B[x],
universe: Type,
bag: bag(T)
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
concat-lifting-loc-3: concat-lifting-loc-3(f),
nat: ℕ,
le: A ≤ B,
and: P ∧ Q,
less_than': less_than'(a;b),
false: False,
not: ¬A,
implies: P ⇒ Q,
prop: ℙ,
int_seg: {i..j-},
uimplies: b supposing a,
guard: {T},
lelt: i ≤ j < k,
all: ∀x:A. B[x],
decidable: Dec(P),
or: P ∨ Q,
satisfiable_int_formula: satisfiable_int_formula(fmla),
exists: ∃x:A. B[x],
top: Top,
sq_type: SQType(T),
select: L[n],
cons: [a / b],
subtract: n - m,
funtype: funtype(n;A;T),
eq_int: (i =z j),
ifthenelse: if b then t else f fi ,
bfalse: ff
Latex:
\mforall{}[A,B,C,D:Type]. \mforall{}[f:Id {}\mrightarrow{} A {}\mrightarrow{} B {}\mrightarrow{} C {}\mrightarrow{} bag(D)].
(concat-lifting-loc-3(f) \mmember{} Id {}\mrightarrow{} bag(A) {}\mrightarrow{} bag(B) {}\mrightarrow{} bag(C) {}\mrightarrow{} bag(D))
Date html generated:
2016_05_17-AM-09_16_03
Last ObjectModification:
2016_01_17-PM-11_15_09
Theory : classrel!lemmas
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