Nuprl Lemma : qinverse_q
∀[a:ℚ]. (((a + -(a)) = 0 ∈ ℚ) ∧ ((-(a) + a) = 0 ∈ ℚ))
Proof
Definitions occuring in Statement :
qmul: r * s,
qadd: r + s,
rationals: ℚ,
uall: ∀[x:A]. B[x],
and: P ∧ Q,
minus: -n,
natural_number: $n,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
subtype_rel: A ⊆r B,
guard: {T},
uimplies: b supposing a,
qadd_grp: <ℚ+>,
grp_car: |g|,
pi1: fst(t),
grp_op: *,
pi2: snd(t),
grp_inv: ~,
grp_id: e,
infix_ap: x f y
Lemmas referenced :
grp_inverse,
qadd_grp_wf,
grp_subtype_igrp,
abgrp_subtype_grp,
subtype_rel_transitivity,
abgrp_wf,
grp_wf,
igrp_wf
Rules used in proof :
cut,
introduction,
extract_by_obid,
sqequalHypSubstitution,
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isectElimination,
thin,
hypothesis,
applyEquality,
instantiate,
independent_isectElimination,
sqequalRule
Latex:
\mforall{}[a:\mBbbQ{}]. (((a + -(a)) = 0) \mwedge{} ((-(a) + a) = 0))
Date html generated:
2020_05_20-AM-09_13_48
Last ObjectModification:
2020_01_25-AM-09_03_20
Theory : rationals
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