Nuprl Lemma : gcd-reduce-prime
∀p,q:ℤ. ∃x,y:ℤ. (((x * p) + (y * q)) = 1 ∈ ℤ) supposing prime(p) ∧ (¬(p | q))
Proof
Definitions occuring in Statement :
prime: prime(a),
divides: b | a,
uimplies: b supposing a,
all: ∀x:A. B[x],
exists: ∃x:A. B[x],
not: ¬A,
and: P ∧ Q,
multiply: n * m,
add: n + m,
natural_number: $n,
int: ℤ,
equal: s = t ∈ T
Definitions unfolded in proof :
all: ∀x:A. B[x],
member: t ∈ T,
uimplies: b supposing a,
prop: ℙ,
and: P ∧ Q,
uall: ∀[x:A]. B[x],
implies: P ⇒ Q,
iff: P ⇐⇒ Q,
rev_implies: P ⇐ Q
Lemmas referenced :
gcd-reduce-coprime,
prime_wf,
not_wf,
divides_wf,
coprime_iff_ndivides
Rules used in proof :
cut,
introduction,
extract_by_obid,
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
lambdaFormation,
hypothesis,
sqequalHypSubstitution,
dependent_functionElimination,
thin,
hypothesisEquality,
isect_memberFormation,
independent_isectElimination,
productEquality,
isectElimination,
intEquality,
productElimination,
independent_functionElimination
Latex:
\mforall{}p,q:\mBbbZ{}. \mexists{}x,y:\mBbbZ{}. (((x * p) + (y * q)) = 1) supposing prime(p) \mwedge{} (\mneg{}(p | q))
Date html generated:
2018_05_21-PM-00_59_25
Last ObjectModification:
2018_05_19-AM-06_35_33
Theory : num_thy_1
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