Nuprl Lemma : sqntype_product
∀[T,S:Type]. ∀[n:ℕ]. (sqntype(n;T × S)) supposing (sqntype(n;S) and sqntype(n;T))
Proof
Definitions occuring in Statement :
sqntype: sqntype(n;T),
nat: ℕ,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
product: x:A × B[x],
universe: Type
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
sqntype: sqntype(n;T),
all: ∀x:A. B[x],
implies: P ⇒ Q,
prop: ℙ,
nat: ℕ,
le: A ≤ B,
and: P ∧ Q,
less_than': less_than'(a;b),
false: False,
not: ¬A
Lemmas referenced :
sqn+1type_product,
equal-wf-base,
base_wf,
sqntype_wf,
nat_wf,
sqequal_n_add,
false_wf,
le_wf
Rules used in proof :
cut,
introduction,
extract_by_obid,
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
hypothesis,
sqequalHypSubstitution,
isectElimination,
thin,
hypothesisEquality,
independent_isectElimination,
lambdaFormation,
dependent_functionElimination,
independent_functionElimination,
productEquality,
universeEquality,
Error :axiomSqequalN,
dependent_set_memberEquality,
natural_numberEquality,
sqequalRule,
independent_pairFormation
Latex:
\mforall{}[T,S:Type]. \mforall{}[n:\mBbbN{}]. (sqntype(n;T \mtimes{} S)) supposing (sqntype(n;S) and sqntype(n;T))
Date html generated:
2019_06_20-AM-11_34_04
Last ObjectModification:
2018_08_17-PM-03_51_00
Theory : int_1
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