Nuprl Lemma : rank-zero
pi-rank(pizero()) = 0 ∈ ℕ
Proof
Definitions occuring in Statement :
pi-rank: pi-rank(p),
pizero: pizero(),
nat: ℕ,
natural_number: $n,
equal: s = t ∈ T
Lemmas :
le_weakening,
le_wf
Latex:
pi-rank(pizero()) = 0
Date html generated:
2015_07_23-AM-11_33_07
Last ObjectModification:
2015_01_29-AM-00_54_03
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