Nuprl Lemma : free-append-assoc
∀[X:Type]. ∀[x,y,z:free-word(X)]. (x + y + z = x + y + z ∈ free-word(X))
Proof
Definitions occuring in Statement :
free-append: w + w',
free-word: free-word(X),
uall: ∀[x:A]. B[x],
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
free-append: w + w',
top: Top
Lemmas referenced :
free-append_wf,
append_assoc
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
cumulativity,
hypothesisEquality,
because_Cache,
hypothesis,
sqequalRule,
isect_memberEquality,
axiomEquality,
voidElimination,
voidEquality
Latex:
\mforall{}[X:Type]. \mforall{}[x,y,z:free-word(X)]. (x + y + z = x + y + z)
Date html generated:
2017_01_19-PM-02_50_18
Last ObjectModification:
2017_01_14-PM-07_34_27
Theory : free!groups
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