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
Definitions unfolded in proof :
pi-rank: pi-rank(p),
pizero: pizero(),
pi_term_ind: pi_term_ind(v;zero;pre,body,rec1....;left,right,rec2,rec3....;left,right,rec4,rec5....;body,rec6....;name,body,rec7....),
all: ∀x:A. B[x],
member: t ∈ T,
decidable: Dec(P),
or: P ∨ Q,
uall: ∀[x:A]. B[x],
uimplies: b supposing a,
satisfiable_int_formula: satisfiable_int_formula(fmla),
exists: ∃x:A. B[x],
top: Top,
prop: ℙ,
false: False,
nat: ℕ,
le: A ≤ B,
and: P ∧ Q,
less_than': less_than'(a;b),
not: ¬A,
implies: P ⇒ Q
Latex:
pi-rank(pizero()) = 0
Date html generated:
2016_05_17-AM-11_23_38
Last ObjectModification:
2016_01_18-AM-07_48_36
Theory : event-logic-applications
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