Nuprl Lemma : list_accum_equality

[T,A,B,C:Type]. ∀[f:A ⟶ T ⟶ A]. ∀[g:B ⟶ T ⟶ B]. ∀[F:A ⟶ C]. ∀[G:B ⟶ C].
  ∀[L:T List]. ∀[a0:A]. ∀[b0:B].
    F[accumulate (with value and list item x):
       f[a;x]
      over list:
        L
      with starting value:
       a0)]
    G[accumulate (with value and list item x):
         g[b;x]
        over list:
          L
        with starting value:
         b0)]
    ∈ 
    supposing F[a0] G[b0] ∈ 
  supposing ∀a:A. ∀b:B. ∀x:T.  (((F a) (G b) ∈ C)  (F[f[a;x]] G[g[b;x]] ∈ C))


Proof




Definitions occuring in Statement :  list_accum: list_accum list: List uimplies: supposing a uall: [x:A]. B[x] so_apply: x[s1;s2] so_apply: x[s] all: x:A. B[x] implies:  Q apply: a function: x:A ⟶ B[x] universe: Type equal: t ∈ T
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T uimplies: supposing a so_lambda: λ2x.t[x] so_apply: x[s] prop: so_lambda: λ2y.t[x; y] so_apply: x[s1;s2] implies:  Q all: x:A. B[x] top: Top
Lemmas referenced :  list_induction uall_wf isect_wf equal_wf list_accum_wf list_wf list_accum_nil_lemma list_accum_cons_lemma all_wf
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut thin extract_by_obid sqequalHypSubstitution isectElimination hypothesisEquality sqequalRule lambdaEquality because_Cache cumulativity applyEquality functionExtensionality hypothesis independent_functionElimination dependent_functionElimination isect_memberEquality voidElimination voidEquality axiomEquality equalityTransitivity equalitySymmetry lambdaFormation rename independent_isectElimination functionEquality universeEquality

Latex:
\mforall{}[T,A,B,C:Type].  \mforall{}[f:A  {}\mrightarrow{}  T  {}\mrightarrow{}  A].  \mforall{}[g:B  {}\mrightarrow{}  T  {}\mrightarrow{}  B].  \mforall{}[F:A  {}\mrightarrow{}  C].  \mforall{}[G:B  {}\mrightarrow{}  C].
    \mforall{}[L:T  List].  \mforall{}[a0:A].  \mforall{}[b0:B].
        F[accumulate  (with  value  a  and  list  item  x):
              f[a;x]
            over  list:
                L
            with  starting  value:
              a0)]
        =  G[accumulate  (with  value  b  and  list  item  x):
                  g[b;x]
                over  list:
                    L
                with  starting  value:
                  b0)] 
        supposing  F[a0]  =  G[b0] 
    supposing  \mforall{}a:A.  \mforall{}b:B.  \mforall{}x:T.    (((F  a)  =  (G  b))  {}\mRightarrow{}  (F[f[a;x]]  =  G[g[b;x]]))



Date html generated: 2017_04_17-AM-07_38_32
Last ObjectModification: 2017_02_27-PM-04_12_54

Theory : list_1


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