Nuprl Lemma : qadd-non-neg
∀[a,b:ℚ]. (0 ≤ (a + b)) supposing ((0 ≤ b) and (0 ≤ a))
Proof
Definitions occuring in Statement :
qle: r ≤ s,
qadd: r + s,
rationals: ℚ,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
natural_number: $n
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
subtype_rel: A ⊆r B,
guard: {T},
uiff: uiff(P;Q),
and: P ∧ Q,
implies: P ⇒ Q,
prop: ℙ,
true: True,
squash: ↓T,
iff: P ⇐⇒ Q
Lemmas referenced :
iff_weakening_equal,
mon_ident_q,
qadd_comm_q,
true_wf,
squash_wf,
qle_transitivity_qorder,
rationals_wf,
qle_wf,
qadd_wf,
int-subtype-rationals,
qle_witness,
qadd_preserves_qle
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
lemma_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
natural_numberEquality,
hypothesis,
applyEquality,
because_Cache,
sqequalRule,
hypothesisEquality,
productElimination,
independent_isectElimination,
independent_functionElimination,
isect_memberEquality,
equalityTransitivity,
equalitySymmetry,
lambdaEquality,
imageElimination,
imageMemberEquality,
baseClosed,
universeEquality
Latex:
\mforall{}[a,b:\mBbbQ{}]. (0 \mleq{} (a + b)) supposing ((0 \mleq{} b) and (0 \mleq{} a))
Date html generated:
2016_05_15-PM-11_04_52
Last ObjectModification:
2016_01_16-PM-09_28_05
Theory : rationals
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