Nuprl Lemma : bm_compare_antisym_le
∀[K:Type]. ∀[compare:bm_compare(K)]. ∀[k1,k2:K]. ((0 ≤ (compare k1 k2)) ⇒ (0 ≤ (compare k2 k1)) ⇒ (k1 = k2 ∈ K))
Proof
Definitions occuring in Statement :
bm_compare: bm_compare(K),
uall: ∀[x:A]. B[x],
le: A ≤ B,
implies: P ⇒ Q,
apply: f a,
natural_number: $n,
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
implies: P ⇒ Q,
bm_compare: bm_compare(K),
and: P ∧ Q,
prop: ℙ,
guard: {T},
anti_sym: AntiSym(T;x,y.R[x; y]),
all: ∀x:A. B[x]
Latex:
\mforall{}[K:Type]. \mforall{}[compare:bm\_compare(K)]. \mforall{}[k1,k2:K].
((0 \mleq{} (compare k1 k2)) {}\mRightarrow{} (0 \mleq{} (compare k2 k1)) {}\mRightarrow{} (k1 = k2))
Date html generated:
2016_05_17-PM-01_40_42
Last ObjectModification:
2015_12_28-PM-08_09_25
Theory : binary-map
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