Nuprl Lemma : arith-example2

`∀f:ℤ ⟶ ℤ. ∀a,b:ℤ.`
`  ((a = b ∈ ℤ)`
`  `` (∀x,y,z:ℤ.`
`        (((f a) ≤ x)`
`        `` (x ≤ (f b))`
`        `` (((x - 1) ≤ y) ∧ (y ≤ ((f b) + 1)))`
`        `` ((((f b) - 1) ≤ z) ∧ (z ≤ (x + 1)))`
`        `` (((y - 1) ≤ z) ∧ (z ≤ (y + 1)))`
`        `` (((x = y ∈ ℤ) ∨ (x = z ∈ ℤ)) ∨ (y = z ∈ ℤ)))))`

Proof

Definitions occuring in Statement :  le: `A ≤ B` all: `∀x:A. B[x]` implies: `P `` Q` or: `P ∨ Q` and: `P ∧ Q` apply: `f a` function: `x:A ⟶ B[x]` subtract: `n - m` add: `n + m` natural_number: `\$n` int: `ℤ` equal: `s = t ∈ T`
Definitions unfolded in proof :  all: `∀x:A. B[x]` implies: `P `` Q` and: `P ∧ Q` member: `t ∈ T` prop: `ℙ` uall: `∀[x:A]. B[x]` subtype_rel: `A ⊆r B` or: `P ∨ Q` le: `A ≤ B` uimplies: `b supposing a` subtract: `n - m` top: `Top` uiff: `uiff(P;Q)` not: `¬A` less_than': `less_than'(a;b)` true: `True` false: `False` iff: `P `⇐⇒` Q` decidable: `Dec(P)` rev_implies: `P `` Q` guard: `{T}`
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity lambdaFormation sqequalHypSubstitution productElimination thin productEquality cut introduction extract_by_obid isectElimination hypothesisEquality natural_numberEquality hypothesis addEquality applyEquality functionExtensionality intEquality sqequalRule functionEquality because_Cache unionElimination independent_isectElimination lambdaEquality isect_memberEquality voidElimination voidEquality minusEquality independent_functionElimination equalitySymmetry dependent_set_memberEquality independent_pairFormation applyLambdaEquality setElimination rename equalityTransitivity dependent_functionElimination inlFormation inrFormation addLevel orFunctionality

Latex:
\mforall{}f:\mBbbZ{}  {}\mrightarrow{}  \mBbbZ{}.  \mforall{}a,b:\mBbbZ{}.
((a  =  b)
{}\mRightarrow{}  (\mforall{}x,y,z:\mBbbZ{}.
(((f  a)  \mleq{}  x)
{}\mRightarrow{}  (x  \mleq{}  (f  b))
{}\mRightarrow{}  (((x  -  1)  \mleq{}  y)  \mwedge{}  (y  \mleq{}  ((f  b)  +  1)))
{}\mRightarrow{}  ((((f  b)  -  1)  \mleq{}  z)  \mwedge{}  (z  \mleq{}  (x  +  1)))
{}\mRightarrow{}  (((y  -  1)  \mleq{}  z)  \mwedge{}  (z  \mleq{}  (y  +  1)))
{}\mRightarrow{}  (((x  =  y)  \mvee{}  (x  =  z))  \mvee{}  (y  =  z)))))

Date html generated: 2017_04_14-AM-07_16_31
Last ObjectModification: 2017_02_27-PM-02_51_51

Theory : arithmetic

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