Nuprl Lemma : module_plus_assoc

[A:Rng]. ∀[m:A-Module]. ∀[x,y,z:m.car].  ((x m.plus (y m.plus z)) ((x m.plus y) m.plus z) ∈ m.car)


Proof




Definitions occuring in Statement :  module: A-Module alg_plus: a.plus alg_car: a.car uall: [x:A]. B[x] infix_ap: y equal: t ∈ T rng: Rng
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T all: x:A. B[x] rng: Rng and: P ∧ Q module: A-Module guard: {T} group_p: IsGroup(T;op;id;inv) monoid_p: IsMonoid(T;op;id) assoc: Assoc(T;op)
Lemmas referenced :  module_properties alg_car_wf rng_car_wf module_wf rng_wf
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut lemma_by_obid sqequalHypSubstitution dependent_functionElimination thin setElimination rename hypothesisEquality hypothesis productElimination isectElimination sqequalRule isect_memberEquality axiomEquality because_Cache

Latex:
\mforall{}[A:Rng].  \mforall{}[m:A-Module].  \mforall{}[x,y,z:m.car].    ((x  m.plus  (y  m.plus  z))  =  ((x  m.plus  y)  m.plus  z))



Date html generated: 2016_05_16-AM-07_26_31
Last ObjectModification: 2015_12_28-PM-05_08_18

Theory : algebras_1


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