Nuprl Lemma : decide-isatom2-if-has-value

t,a,b:Base.  ((t)↓  ((isatom2(t;a;b) a) ∨ (isatom2(t;a;b) b)))


Proof




Definitions occuring in Statement :  has-value: (a)↓ isatom2: isatom2(z;a;b) all: x:A. B[x] implies:  Q or: P ∨ Q base: Base sqequal: t
Definitions unfolded in proof :  all: x:A. B[x] implies:  Q has-value: (a)↓ member: t ∈ T uall: [x:A]. B[x] or: P ∨ Q top: Top guard: {T} prop:
Lemmas referenced :  base_wf top_wf is-exception_wf has-value_wf_base
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity lambdaFormation isatom2Cases divergentSqle hypothesis cut lemma_by_obid sqequalHypSubstitution isectElimination thin baseClosed hypothesisEquality sqequalRule isAtom2ReduceTrue equalityTransitivity equalitySymmetry because_Cache inlFormation sqequalIntensionalEquality isect_memberFormation introduction sqequalAxiom isect_memberEquality voidElimination voidEquality inrFormation

Latex:
\mforall{}t,a,b:Base.    ((t)\mdownarrow{}  {}\mRightarrow{}  ((isatom2(t;a;b)  \msim{}  a)  \mvee{}  (isatom2(t;a;b)  \msim{}  b)))



Date html generated: 2016_05_13-PM-03_22_32
Last ObjectModification: 2016_01_14-PM-06_46_34

Theory : call!by!value_1


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