Nuprl Lemma : add-is-int-iff
∀[a,b:Base]. uiff(a + b ∈ ℤ;(a ∈ ℤ) ∧ (b ∈ ℤ))
Proof
Definitions occuring in Statement :
uiff: uiff(P;Q),
uall: ∀[x:A]. B[x],
and: P ∧ Q,
member: t ∈ T,
add: n + m,
int: ℤ,
base: Base
Definitions unfolded in proof :
prop: ℙ,
uimplies: b supposing a,
and: P ∧ Q,
uiff: uiff(P;Q),
member: t ∈ T,
uall: ∀[x:A]. B[x]
Lemmas referenced :
base_wf,
equal-wf-base
Rules used in proof :
isect_memberEquality,
productEquality,
because_Cache,
hypothesisEquality,
baseClosed,
closedConclusion,
baseApply,
intEquality,
isectElimination,
extract_by_obid,
equalitySymmetry,
hypothesis,
equalityTransitivity,
axiomEquality,
independent_pairEquality,
thin,
productElimination,
sqequalHypSubstitution,
sqequalRule,
independent_pairFormation,
cut,
introduction,
isect_memberFormation,
sqequalReflexivity,
computationStep,
sqequalTransitivity,
sqequalSubstitution,
callbyvalueAdd,
callbyvalueInt,
addEquality
Latex:
\mforall{}[a,b:Base]. uiff(a + b \mmember{} \mBbbZ{};(a \mmember{} \mBbbZ{}) \mwedge{} (b \mmember{} \mBbbZ{}))
Date html generated:
2019_06_20-AM-11_21_55
Last ObjectModification:
2018_10_15-AM-11_17_52
Theory : arithmetic
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