Nuprl Lemma : zero-vector-add-left
∀[i:Type]. ∀[r:Rng]. ∀[a:i ⟶ |r|]. ((0 + a) = a ∈ (i ⟶ |r|))
Proof
Definitions occuring in Statement :
zero-vector: 0,
vector-add: (a + b),
uall: ∀[x:A]. B[x],
function: x:A ⟶ B[x],
universe: Type,
equal: s = t ∈ T,
rng: Rng,
rng_car: |r|
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
zero-vector: 0,
vector-add: (a + b),
and: P ∧ Q,
rng: Rng
Lemmas referenced :
rng_plus_zero,
rng_car_wf,
rng_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
functionExtensionality,
sqequalRule,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
hypothesisEquality,
applyEquality,
productElimination,
hypothesis,
functionEquality,
setElimination,
rename,
isect_memberEquality,
axiomEquality,
because_Cache,
universeEquality
Latex:
\mforall{}[i:Type]. \mforall{}[r:Rng]. \mforall{}[a:i {}\mrightarrow{} |r|]. ((0 + a) = a)
Date html generated:
2018_05_21-PM-09_40_48
Last ObjectModification:
2018_05_19-PM-04_33_09
Theory : matrices
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