### Nuprl Lemma : rng_plus_inv

`∀[r:Rng]. ∀[a:|r|].  (((a +r (-r a)) = 0 ∈ |r|) ∧ (((-r a) +r a) = 0 ∈ |r|))`

Proof

Definitions occuring in Statement :  rng: `Rng` rng_minus: `-r` rng_zero: `0` rng_plus: `+r` rng_car: `|r|` uall: `∀[x:A]. B[x]` infix_ap: `x f y` and: `P ∧ Q` apply: `f a` equal: `s = t ∈ T`
Definitions unfolded in proof :  uall: `∀[x:A]. B[x]` member: `t ∈ T` subtype_rel: `A ⊆r B` add_grp_of_rng: `r↓+gp` grp_car: `|g|` pi1: `fst(t)` grp_op: `*` pi2: `snd(t)` grp_inv: `~` grp_id: `e` and: `P ∧ Q` rng: `Rng`
Lemmas referenced :  grp_inverse add_grp_of_rng_wf_a grp_subtype_igrp rng_car_wf rng_wf
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut lemma_by_obid sqequalHypSubstitution isectElimination thin hypothesisEquality hypothesis applyEquality sqequalRule isect_memberEquality productElimination independent_pairEquality axiomEquality setElimination rename

Latex:
\mforall{}[r:Rng].  \mforall{}[a:|r|].    (((a  +r  (-r  a))  =  0)  \mwedge{}  (((-r  a)  +r  a)  =  0))

Date html generated: 2016_05_15-PM-00_21_19
Last ObjectModification: 2015_12_27-AM-00_02_20

Theory : rings_1

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