Nuprl Lemma : add-inverse-unique
∀[x,y:ℤ]. (((x + y) = 0 ∈ ℤ) ⇒ (y = (-x) ∈ ℤ))
Proof
Definitions occuring in Statement :
uall: ∀[x:A]. B[x],
implies: P ⇒ Q,
add: n + m,
minus: -n,
natural_number: $n,
int: ℤ,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
implies: P ⇒ Q,
prop: ℙ,
subtype_rel: A ⊆r B,
top: Top
Lemmas referenced :
equal-wf-base,
int_subtype_base,
add-associates,
add-inverse2,
add-zero,
zero-add
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
lambdaFormation,
hypothesis,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
intEquality,
sqequalRule,
baseApply,
closedConclusion,
baseClosed,
hypothesisEquality,
applyEquality,
because_Cache,
lambdaEquality,
dependent_functionElimination,
axiomEquality,
isect_memberEquality,
addEquality,
minusEquality,
voidElimination,
voidEquality
Latex:
\mforall{}[x,y:\mBbbZ{}]. (((x + y) = 0) {}\mRightarrow{} (y = (-x)))
Date html generated:
2017_04_14-AM-07_16_16
Last ObjectModification:
2017_02_27-PM-02_51_07
Theory : arithmetic
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