Nuprl Lemma : fun_exp_com

[T:Type]. ∀[n,m:ℕ]. ∀[f:T ⟶ T]. ∀[x:T].  ((f^m (f^n x)) (f^n (f^m x)) ∈ T)


Proof




Definitions occuring in Statement :  fun_exp: f^n nat: uall: [x:A]. B[x] apply: a function: x:A ⟶ B[x] universe: Type equal: t ∈ T
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T squash: T prop: true: True subtype_rel: A ⊆B uimplies: supposing a guard: {T} iff: ⇐⇒ Q and: P ∧ Q rev_implies:  Q implies:  Q nat: top: Top all: x:A. B[x] sq_stable: SqStable(P)
Lemmas referenced :  equal_wf squash_wf true_wf fun_exp_add_apply fun_exp_wf iff_weakening_equal add-commutes add_nat_wf nat_wf sq_stable__le le_wf
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut applyEquality thin lambdaEquality sqequalHypSubstitution imageElimination extract_by_obid isectElimination hypothesisEquality equalityTransitivity hypothesis equalitySymmetry universeEquality because_Cache cumulativity functionExtensionality natural_numberEquality sqequalRule imageMemberEquality baseClosed independent_isectElimination productElimination independent_functionElimination setElimination rename isect_memberEquality voidElimination voidEquality dependent_set_memberEquality addEquality lambdaFormation dependent_functionElimination axiomEquality functionEquality

Latex:
\mforall{}[T:Type].  \mforall{}[n,m:\mBbbN{}].  \mforall{}[f:T  {}\mrightarrow{}  T].  \mforall{}[x:T].    ((f\^{}m  (f\^{}n  x))  =  (f\^{}n  (f\^{}m  x)))



Date html generated: 2017_04_14-AM-07_34_45
Last ObjectModification: 2017_02_27-PM-03_07_38

Theory : fun_1


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