Nuprl Lemma : comp_assoc
∀[A,B,C,D:Type]. ∀[f:A ⟶ B]. ∀[g:B ⟶ C]. ∀[h:C ⟶ D]. ((h o (g o f)) = ((h o g) o f) ∈ (A ⟶ D))
Proof
Definitions occuring in Statement :
compose: f o g,
uall: ∀[x:A]. B[x],
function: x:A ⟶ B[x],
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
compose: f o g,
member: t ∈ T,
uall: ∀[x:A]. B[x]
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
sqequalRule,
lambdaEquality,
applyEquality,
hypothesisEquality,
Error :functionIsType,
Error :universeIsType,
functionEquality,
because_Cache,
Error :inhabitedIsType,
universeEquality,
Error :isect_memberFormation_alt,
introduction,
cut,
hypothesis,
sqequalHypSubstitution,
isect_memberEquality,
isectElimination,
thin,
axiomEquality
Latex:
\mforall{}[A,B,C,D:Type]. \mforall{}[f:A {}\mrightarrow{} B]. \mforall{}[g:B {}\mrightarrow{} C]. \mforall{}[h:C {}\mrightarrow{} D]. ((h o (g o f)) = ((h o g) o f))
Date html generated:
2019_06_20-PM-00_26_16
Last ObjectModification:
2018_09_26-AM-11_50_35
Theory : fun_1
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