Nuprl Lemma : add_eqmod_zero
∀m,x,y:ℤ. ((x ≡ 0 mod m) ⇒ (y ≡ 0 mod m) ⇒ ((x + y) ≡ 0 mod m))
Proof
Definitions occuring in Statement :
eqmod: a ≡ b mod m,
all: ∀x:A. B[x],
implies: P ⇒ Q,
add: n + m,
natural_number: $n,
int: ℤ
Definitions unfolded in proof :
all: ∀x:A. B[x],
implies: P ⇒ Q,
member: t ∈ T,
prop: ℙ,
uall: ∀[x:A]. B[x],
top: Top,
squash: ↓T,
true: True
Lemmas referenced :
eqmod_wf,
add-commutes,
zero-add,
add_functionality_wrt_eqmod
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
lambdaFormation,
cut,
introduction,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
hypothesisEquality,
natural_numberEquality,
hypothesis,
intEquality,
sqequalRule,
isect_memberEquality,
voidElimination,
voidEquality,
addEquality,
because_Cache,
applyEquality,
lambdaEquality,
imageElimination,
imageMemberEquality,
baseClosed,
equalitySymmetry,
hyp_replacement,
Error :applyLambdaEquality,
dependent_functionElimination,
independent_functionElimination
Latex:
\mforall{}m,x,y:\mBbbZ{}. ((x \mequiv{} 0 mod m) {}\mRightarrow{} (y \mequiv{} 0 mod m) {}\mRightarrow{} ((x + y) \mequiv{} 0 mod m))
Date html generated:
2016_10_21-AM-11_09_07
Last ObjectModification:
2016_07_12-AM-06_01_37
Theory : num_thy_1
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