Nuprl Lemma : gcd-lcm-absorption
∀[n,m:ℕ]. (gcd(n;lcm(n;m)) = n ∈ ℤ)
Proof
Definitions occuring in Statement :
lcm: lcm(a;b),
gcd: gcd(a;b),
nat: ℕ,
uall: ∀[x:A]. B[x],
int: ℤ,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
and: P ∧ Q,
cand: A c∧ B,
all: ∀x:A. B[x],
nat: ℕ,
implies: P ⇒ Q,
prop: ℙ
Lemmas referenced :
nat_wf,
lcm_wf_nat,
divides_reflexivity,
divides_wf,
lcm_wf,
gcd-unique-nat,
lcm-is-lcm-nat
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
equalitySymmetry,
hypothesis,
lemma_by_obid,
sqequalRule,
sqequalHypSubstitution,
isect_memberEquality,
isectElimination,
thin,
hypothesisEquality,
axiomEquality,
because_Cache,
dependent_functionElimination,
setElimination,
rename,
independent_pairFormation,
productElimination,
lambdaFormation,
intEquality,
independent_functionElimination
Latex:
\mforall{}[n,m:\mBbbN{}]. (gcd(n;lcm(n;m)) = n)
Date html generated:
2016_05_14-PM-09_25_48
Last ObjectModification:
2015_12_26-PM-08_02_52
Theory : num_thy_1
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