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0c1a124704
Division Theorem. For every natural number `x` and positive natural
number `n`, there is a unique decomposition:
`x = q*n + r`
with `q`,`r` natural and `r` < `n`.
`q` is the quotient when dividing `x` by `n`
`r` is the remainder when dividing `x` by `n`.
This commit adds a proof for this fact, in case
we want to reason about modular arithmetic (for example, when dealing
with binary representations). A future, more systematic, development could
perhaps follow: @clayrat 's (idris1) port of Coq's binary arithmetics:
https://github.com/sbp/idris-bi/blob/master/src/Data/Bin/DivMod.idr
https://github.com/sbp/idris-bi/blob/master/src/Data/Biz/DivMod.idr
https://github.com/sbp/idris-bi/blob/master/src/Data/BizMod2/DivMod.idr
In the process, it bulks up the stdlib with:
+ a generic PreorderReasoning module for arbitrary preorders,
analogous for the equational reasoning module
+ some missing facts about Nat operations.
+ Refactor some Nat order properties using a 'reflect' function
Co-authored-by: Ohad Kammar <ohad.kammar@ed.ac.uk>
Co-authored-by: G. Allais <guillaume.allais@ens-lyon.org>
77 lines
2.6 KiB
Idris
77 lines
2.6 KiB
Idris
module Data.Nat.Equational
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import Data.Nat
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%default total
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||| Subtract a number from both sides of an equation.
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||| Due to partial nature of subtraction in natural numbers, an equation of
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||| special form is required in order for subtraction to be total.
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export
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subtractEqLeft : (a : Nat) -> {b, c : Nat} -> a + b = a + c -> b = c
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subtractEqLeft 0 prf = prf
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subtractEqLeft (S k) prf = subtractEqLeft k $ succInjective (k + b) (k + c) prf
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||| Subtract a number from both sides of an equation.
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||| Due to partial nature of subtraction in natural numbers, an equation of
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export
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subtractEqRight : {a, b : Nat} -> (c : Nat) -> a + c = b + c -> a = b
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subtractEqRight c prf =
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subtractEqLeft c $
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rewrite plusCommutative c a in
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rewrite plusCommutative c b in
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prf
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||| Add a number to both sides of an inequality
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export
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plusLteLeft : (a : Nat) -> {b, c : Nat} -> LTE b c -> LTE (a + b) (a + c)
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plusLteLeft 0 bLTEc = bLTEc
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plusLteLeft (S k) bLTEc = LTESucc $ plusLteLeft k bLTEc
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||| Add a number to both sides of an inequality
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export
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plusLteRight : {a, b : Nat} -> (c : Nat) -> LTE a b -> LTE (a + c) (b + c)
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plusLteRight c aLTEb =
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rewrite plusCommutative a c in
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rewrite plusCommutative b c in
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plusLteLeft c aLTEb
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||| Only 0 is lte 0
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||| Useful when the argument is an open term
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export
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lteZeroIsZero : a `LTE` 0 -> a = 0
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lteZeroIsZero LTEZero = Refl
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||| Subtract a number from both sides of an inequality.
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||| Due to partial nature of subtraction, we require an inequality of special form.
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export
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subtractLteLeft : (a : Nat) -> {b, c : Nat} -> LTE (a + b) (a + c) -> LTE b c
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subtractLteLeft 0 abLTEac = abLTEac
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subtractLteLeft (S k) abLTEac = subtractLteLeft k $ fromLteSucc abLTEac
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||| Subtract a number from both sides of an inequality.
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||| Due to partial nature of subtraction, we require an inequality of special form.
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export
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subtractLteRight : {a, b : Nat} -> (c : Nat) -> LTE (a + c) (b + c) -> LTE a b
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subtractLteRight c acLTEbc =
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subtractLteLeft c $
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rewrite plusCommutative c a in
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rewrite plusCommutative c b in
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acLTEbc
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||| If one of the factors of a product is greater than 0, then the other factor
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||| is less than or equal to the product..
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export
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rightFactorLteProduct : (a, b : Nat) -> LTE b (S a * b)
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rightFactorLteProduct a b = lteAddRight b
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||| If one of the factors of a product is greater than 0, then the other factor
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export
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leftFactorLteProduct : (a, b : Nat) -> LTE a (a * S b)
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leftFactorLteProduct a b =
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rewrite multRightSuccPlus a b in
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lteAddRight a
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