Nuprl Lemma : gcd_mul
∀a,b,n:ℤ. ((n * gcd(a;b)) ~ gcd(n * a;n * b))
Proof
Definitions occuring in Statement :
assoced: a ~ b,
gcd: gcd(a;b),
all: ∀x:A. B[x],
multiply: n * m,
int: ℤ
Definitions unfolded in proof :
all: ∀x:A. B[x],
member: t ∈ T,
implies: P ⇒ Q,
prop: ℙ,
uall: ∀[x:A]. B[x]
Lemmas referenced :
istype-int,
gcd_wf,
gcd_sat_pred,
gcd_p_wf,
gcd_p_mul,
gcd_unique
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
Error :lambdaFormation_alt,
Error :inhabitedIsType,
hypothesisEquality,
cut,
introduction,
extract_by_obid,
hypothesis,
sqequalHypSubstitution,
dependent_functionElimination,
thin,
equalitySymmetry,
hyp_replacement,
applyLambdaEquality,
isectElimination,
multiplyEquality,
Error :equalityIsType1,
equalityTransitivity,
independent_functionElimination
Latex:
\mforall{}a,b,n:\mBbbZ{}. ((n * gcd(a;b)) \msim{} gcd(n * a;n * b))
Date html generated:
2019_06_20-PM-02_22_30
Last ObjectModification:
2018_10_03-AM-00_12_23
Theory : num_thy_1
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