Nuprl Lemma : equality-test
∀[A,B,C:Type]. ((A = B ∈ Type) ⇒ (C = B ∈ Type) ⇒ (∀[x,y,z:A]. ((x = y ∈ A) ⇒ (z = y ∈ A) ⇒ (x = z ∈ C))))
Proof
Definitions occuring in Statement :
uall: ∀[x:A]. B[x],
implies: P ⇒ Q,
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
implies: P ⇒ Q,
member: t ∈ T,
prop: ℙ,
subtype_rel: A ⊆r B
Lemmas referenced :
equal_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
Error :isect_memberFormation_alt,
Error :lambdaFormation_alt,
Error :universeIsType,
cut,
introduction,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
hypothesisEquality,
hypothesis,
Error :inhabitedIsType,
instantiate,
universeEquality,
equalityTransitivity,
equalitySymmetry,
applyEquality,
sqequalRule,
Error :lambdaEquality_alt,
hyp_replacement
Latex:
\mforall{}[A,B,C:Type]. ((A = B) {}\mRightarrow{} (C = B) {}\mRightarrow{} (\mforall{}[x,y,z:A]. ((x = y) {}\mRightarrow{} (z = y) {}\mRightarrow{} (x = z))))
Date html generated:
2019_06_20-AM-11_18_38
Last ObjectModification:
2018_09_27-PM-05_34_18
Theory : core_2
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