Nuprl Lemma : subtract-is-int-iff

[a,b:Base].  uiff(a b ∈ ℤ;(a ∈ ℤ) ∧ (b ∈ ℤ))


Proof




Definitions occuring in Statement :  uiff: uiff(P;Q) uall: [x:A]. B[x] and: P ∧ Q member: t ∈ T subtract: m int: base: Base
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T subtract: m uiff: uiff(P;Q) and: P ∧ Q uimplies: supposing a prop: implies:  Q iff: ⇐⇒ Q rev_implies:  Q
Lemmas referenced :  minus-is-int-iff add-is-int-iff iff_weakening_uiff iff_transitivity uiff_wf base_wf equal-wf-base
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut sqequalRule sqequalHypSubstitution productElimination thin independent_pairEquality isect_memberEquality isectElimination hypothesisEquality axiomEquality equalityTransitivity hypothesis equalitySymmetry lemma_by_obid intEquality baseApply closedConclusion baseClosed because_Cache productEquality independent_pairFormation instantiate cumulativity addLevel independent_isectElimination independent_functionElimination lambdaFormation

Latex:
\mforall{}[a,b:Base].    uiff(a  -  b  \mmember{}  \mBbbZ{};(a  \mmember{}  \mBbbZ{})  \mwedge{}  (b  \mmember{}  \mBbbZ{}))



Date html generated: 2016_05_13-PM-03_28_37
Last ObjectModification: 2016_01_14-PM-06_42_07

Theory : arithmetic


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