Nuprl Lemma : int-rdiv-cancel
∀a:ℤ. ∀n,m:ℕ+. ((r(m * a))/m * n = (r(a))/n ∈ ℝ)
Proof
Definitions occuring in Statement :
int-rdiv: (a)/k1,
int-to-real: r(n),
real: ℝ,
nat_plus: ℕ+,
all: ∀x:A. B[x],
multiply: n * m,
int: ℤ,
equal: s = t ∈ T
Definitions unfolded in proof :
all: ∀x:A. B[x],
member: t ∈ T,
uall: ∀[x:A]. B[x],
int_nzero: ℤ-o,
nat_plus: ℕ+,
subtype_rel: A ⊆r B,
nequal: a ≠ b ∈ T ,
prop: ℙ,
real: ℝ,
squash: ↓T,
int-to-real: r(n),
int-rdiv: (a)/k1,
has-value: (a)↓,
uimplies: b supposing a,
so_lambda: λ2x.t[x],
so_apply: x[s],
decidable: Dec(P),
or: P ∨ Q,
not: ¬A,
implies: P ⇒ Q,
satisfiable_int_formula: satisfiable_int_formula(fmla),
exists: ∃x:A. B[x],
false: False,
sq_type: SQType(T),
guard: {T}
Latex:
\mforall{}a:\mBbbZ{}. \mforall{}n,m:\mBbbN{}\msupplus{}. ((r(m * a))/m * n = (r(a))/n)
Date html generated:
2020_05_20-AM-11_07_31
Last ObjectModification:
2019_12_28-PM-08_14_50
Theory : reals
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