Nuprl Lemma : gcd_is_gcd
∀a,b,c:ℤ. ((c | a) ⇒ (c | b) ⇒ (c | gcd(a;b)))
Proof
Definitions occuring in Statement :
divides: b | a,
gcd: gcd(a;b),
all: ∀x:A. B[x],
implies: P ⇒ Q,
int: ℤ
Definitions unfolded in proof :
all: ∀x:A. B[x],
implies: P ⇒ Q,
member: t ∈ T,
uall: ∀[x:A]. B[x],
prop: ℙ,
gcd_p: GCD(a;b;y),
and: P ∧ Q
Lemmas referenced :
divides_wf,
istype-int,
gcd_wf,
gcd_sat_pred,
gcd_p_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
Error :lambdaFormation_alt,
Error :universeIsType,
cut,
introduction,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
hypothesisEquality,
hypothesis,
Error :inhabitedIsType,
dependent_functionElimination,
equalitySymmetry,
hyp_replacement,
applyLambdaEquality,
Error :equalityIsType1,
equalityTransitivity,
independent_functionElimination,
productElimination,
independent_pairFormation
Latex:
\mforall{}a,b,c:\mBbbZ{}. ((c | a) {}\mRightarrow{} (c | b) {}\mRightarrow{} (c | gcd(a;b)))
Date html generated:
2019_06_20-PM-02_22_09
Last ObjectModification:
2018_10_03-AM-00_12_16
Theory : num_thy_1
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