Nuprl Lemma : comp_id_l
∀[A,B:Type]. ∀[f:A ⟶ B]. ((Id{B} o f) = f ∈ (A ⟶ B))
Proof
Definitions occuring in Statement :
compose: f o g,
tidentity: Id{T},
uall: ∀[x:A]. B[x],
function: x:A ⟶ B[x],
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
tidentity: Id{T},
identity: Id,
compose: f o g,
uall: ∀[x:A]. B[x],
member: t ∈ T
Rules used in proof :
sqequalSubstitution,
sqequalRule,
sqequalReflexivity,
sqequalTransitivity,
computationStep,
Error :isect_memberFormation_alt,
introduction,
cut,
functionExtensionality,
applyEquality,
hypothesisEquality,
hypothesis,
Error :functionIsType,
Error :universeIsType,
sqequalHypSubstitution,
isect_memberEquality,
isectElimination,
thin,
axiomEquality,
functionEquality,
Error :inhabitedIsType,
because_Cache,
universeEquality
Latex:
\mforall{}[A,B:Type]. \mforall{}[f:A {}\mrightarrow{} B]. ((Id\{B\} o f) = f)
Date html generated:
2019_06_20-PM-00_26_17
Last ObjectModification:
2018_09_26-AM-11_50_35
Theory : fun_1
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