Nuprl Lemma : add_cancel_in_eq
∀[a,b,n:ℤ]. a = b ∈ ℤ supposing (a + n) = (b + n) ∈ ℤ
Proof
Definitions occuring in Statement :
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
add: n + m,
int: ℤ,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
prop: ℙ,
subtract: n - m,
subtype_rel: A ⊆r B,
top: Top,
squash: ↓T,
true: True
Lemmas referenced :
true_wf,
squash_wf,
subtract_wf,
add-zero,
zero-mul,
add-mul-special,
minus-one-mul,
add-associates,
equal_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
hypothesis,
lemma_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
intEquality,
addEquality,
hypothesisEquality,
sqequalRule,
isect_memberEquality,
axiomEquality,
because_Cache,
equalityTransitivity,
equalitySymmetry,
applyEquality,
lambdaEquality,
voidElimination,
voidEquality,
minusEquality,
natural_numberEquality,
imageElimination,
imageMemberEquality,
baseClosed
Latex:
\mforall{}[a,b,n:\mBbbZ{}]. a = b supposing (a + n) = (b + n)
Date html generated:
2016_05_13-PM-03_39_39
Last ObjectModification:
2016_01_14-PM-06_38_10
Theory : arithmetic
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