Nuprl Lemma : gcd-unique-nat
∀n,m,g:ℕ. ((((g | n) ∧ (g | m)) ∧ (∀v:ℤ. ((v | n) ⇒ (v | m) ⇒ (v | g)))) ⇒ (g = gcd(n;m) ∈ ℤ))
Proof
Definitions occuring in Statement :
divides: b | a,
gcd: gcd(a;b),
nat: ℕ,
all: ∀x:A. B[x],
implies: P ⇒ Q,
and: P ∧ Q,
int: ℤ,
equal: s = t ∈ T
Definitions unfolded in proof :
all: ∀x:A. B[x],
implies: P ⇒ Q,
and: P ∧ Q,
member: t ∈ T,
nat: ℕ,
uall: ∀[x:A]. B[x],
prop: ℙ,
iff: P ⇐⇒ Q,
so_lambda: λ2x.t[x],
so_apply: x[s],
gcd_p: GCD(a;b;y),
cand: A c∧ B,
guard: {T}
Lemmas referenced :
assoced_nelim,
gcd_wf,
gcd-non-neg,
le_wf,
gcd_unique,
gcd_sat_pred,
and_wf,
divides_wf,
all_wf,
nat_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
lambdaFormation,
cut,
sqequalHypSubstitution,
productElimination,
thin,
lemma_by_obid,
dependent_functionElimination,
hypothesisEquality,
dependent_set_memberEquality,
setElimination,
rename,
hypothesis,
isectElimination,
natural_numberEquality,
independent_functionElimination,
because_Cache,
intEquality,
sqequalRule,
lambdaEquality,
functionEquality,
independent_pairFormation
Latex:
\mforall{}n,m,g:\mBbbN{}. ((((g | n) \mwedge{} (g | m)) \mwedge{} (\mforall{}v:\mBbbZ{}. ((v | n) {}\mRightarrow{} (v | m) {}\mRightarrow{} (v | g)))) {}\mRightarrow{} (g = gcd(n;m)))
Date html generated:
2016_05_14-PM-09_25_24
Last ObjectModification:
2015_12_26-PM-08_03_03
Theory : num_thy_1
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