Nuprl Lemma : lcm-com-nat
∀n,m:ℕ. (lcm(n;m) = lcm(m;n) ∈ ℤ)
Proof
Definitions occuring in Statement :
lcm: lcm(a;b),
nat: ℕ,
all: ∀x:A. B[x],
int: ℤ,
equal: s = t ∈ T
Definitions unfolded in proof :
all: ∀x:A. B[x],
member: t ∈ T,
uall: ∀[x:A]. B[x],
implies: P ⇒ Q,
and: P ∧ Q,
cand: A c∧ B,
prop: ℙ,
nat: ℕ,
guard: {T}
Lemmas referenced :
lcm-is-lcm-nat,
lcm-unique-nat,
lcm_wf_nat,
divides_wf,
nat_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
lambdaFormation,
cut,
lemma_by_obid,
sqequalHypSubstitution,
dependent_functionElimination,
thin,
hypothesisEquality,
because_Cache,
isectElimination,
hypothesis,
independent_functionElimination,
productElimination,
independent_pairFormation,
setElimination,
rename,
intEquality
Latex:
\mforall{}n,m:\mBbbN{}. (lcm(n;m) = lcm(m;n))
Date html generated:
2016_05_14-PM-09_25_35
Last ObjectModification:
2015_12_26-PM-08_02_49
Theory : num_thy_1
Home
Index