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Field_ZF

Wed, 29 Oct 2008 23:46:00 GMT

theory Field_ZF imports Ring_ZF

begin

This theory covers basic facts about fields.

Definition and basic properties

In this section we define what is a field and list the basic properties of fields.

Field is a notrivial commutative ring such that all non-zero elements have an inverse. We define the notion of being a field as a statement about three sets. The first set, denoted K is the carrier of the field. The second set, denoted A represents the additive operation on K (recall that in ZF set theory functions are sets). The third set M represents the multiplicative operation on K.

Definition
IsAfield(K,A,M) \equiv
(IsAring(K,A,M) \wedge (M \text{ is commutative on } K) \wedge
TheNeutralElement(K,A) \neq TheNeutralElement(K,M) \wedge
(\forall a\in K.\ a\neq TheNeutralElement(K,A)\longrightarrow
(\exists b\in K.\ M\langle a,b\rangle = TheNeutralElement(K,M))))

The field0 context extends the ring0 context adding field-related assumptions and notation related to the multiplicative inverse.

Locale field0 = ring0 K +
assumes mult_commute: M \text{ is commutative on } K
assumes not_triv: 0 \neq 1
assumes inv_exists: \forall a\in K.\ a\neq 0 \longrightarrow (\exists b\in K.\ a\cdot b = 1 )
defines K_0 \equiv K-\{0 \}
defines a^{-1} \equiv GroupInv(K_0,restrict(M,K_0\times K_0))(a)

The next lemma assures us that we are talking fields in the field0 context.

lemma (in field0 ) Field_ZF_1_L1:
shows IsAfield(K,A,M) using ringAssum , mult_commute , not_triv , inv_exists , IsAfield_def

We can use theorems proven in the field0 context whenever we talk about a field.

lemma field_field0:
assumes IsAfield(K,A,M) shows field0(K,A,M) using assms , IsAfield_def , field0_axioms.intro , ring0_def , field0_def

Let's have an explicit statement that the multiplication in fields is commutative.

lemma (in field0 ) field_mult_comm:
assumes a\in K, b\in K shows a\cdot b = b\cdot a using mult_commute , assms , IsCommutative_def

Fields do not have zero divisors.

lemma (in field0 ) field_has_no_zero_divs:
shows HasNoZeroDivs(K,A,M)proof

K_0 (the set of nonzero field elements is closed with respect to multiplication.

lemma (in field0 ) Field_ZF_1_L2:
shows K_0 \text{ is closed under } M using Ring_ZF_1_L4 , field_has_no_zero_divs , Ring_ZF_1_L12 , IsOpClosed_def

Any nonzero element has a right inverse that is nonzero.

lemma (in field0 ) Field_ZF_1_L3:
assumes A1: a\in K_0 shows \exists b\in K_0.\ a\cdot b = 1 proof

If we remove zero, the field with multiplication becomes a group and we can use all theorems proven in group0 context.

theorem (in field0 ) Field_ZF_1_L4:
shows \text{IsAgroup}(K_0,restrict(M,K_0\times K_0)), group0(K_0,restrict(M,K_0\times K_0)), 1 = TheNeutralElement(K_0,restrict(M,K_0\times K_0))proof

The inverse of a nonzero field element is nonzero.

lemma (in field0 ) Field_ZF_1_L5:
assumes A1: a\in K, a\neq 0 shows a^{-1} \in K_0, (a^{-1})^2 \in K_0, a^{-1} \in K, a^{-1} \neq 0 proof

The inverse is really the inverse.

lemma (in field0 ) Field_ZF_1_L6:
assumes A1: a\in K, a\neq 0 shows a\cdot a^{-1} = 1 , a^{-1}\cdot a = 1 proof

A lemma with two field elements and cancelling.

lemma (in field0 ) Field_ZF_1_L7:
assumes a\in K, b\in K, b\neq 0 shows a\cdot b\cdot b^{-1} = a, a\cdot b^{-1}\cdot b = a using assms , Field_ZF_1_L5 , Ring_ZF_1_L11 , Field_ZF_1_L6 , Ring_ZF_1_L3

Equations and identities

This section deals with more specialized identities that are true in fields.

a/(a^2) = a.

lemma (in field0 ) Field_ZF_2_L1:
assumes A1: a\in K, a\neq 0 shows a\cdot (a^{-1})^2 = a^{-1}proof

If we multiply two different numbers by a nonzero number, the results will be different.

lemma (in field0 ) Field_ZF_2_L2:
assumes a\in K, b\in K, c\in K, a\neq b, c\neq 0 shows a\cdot c^{-1} \neq b\cdot c^{-1} using assms , field_has_no_zero_divs , Field_ZF_1_L5 , Ring_ZF_1_L12B

We can put a nonzero factor on the other side of non-identity (is this the best way to call it?) changing it to the inverse.

lemma (in field0 ) Field_ZF_2_L3:
assumes A1: a\in K, b\in K, b\neq 0 , c\in K and A2: a\cdot b \neq c shows a \neq c\cdot b^{-1}proof

If if the inverse of b is different than a, then the inverse of a is different than b.

lemma (in field0 ) Field_ZF_2_L4:
assumes a\in K, a\neq 0 and b^{-1} \neq a shows a^{-1} \neq b using assms , Field_ZF_1_L4 , group0_2_L11B

An identity with two field elements, one and an inverse.

lemma (in field0 ) Field_ZF_2_L5:
assumes a\in K, b\in K, b\neq 0 shows (1 + a\cdot b)\cdot b^{-1} = a + b^{-1} using assms , Ring_ZF_1_L4 , Field_ZF_1_L5 , Ring_ZF_1_L2 , ring_oper_distr , Field_ZF_1_L7 , Ring_ZF_1_L3

An identity with three field elements, inverse and cancelling.

lemma (in field0 ) Field_ZF_2_L6:
assumes A1: a\in K, b\in K, b\neq 0 , c\in K shows a\cdot b\cdot (c\cdot b^{-1}) = a\cdot cproof

end

Comments on Field_ZF

AbelianGroup_ZF

Sun, 26 Oct 2008 22:53:00 GMT

theory AbelianGroup_ZF imports Group_ZF

begin

A group is called ``abelian`` if its operation is commutative, i.e. P\langle a,b \rangle = P\langle a,b \rangle for all group elements a,b, where P is the group operation. It is customary to use the additive notation for abelian groups, so this condition is typically written as a+b = b+a. We will be using multiplicative notation though (in which the commutativity condition of the operation is written as a\cdot b = b\cdot a), just to avoid the hassle of changing the notation we used for general groups.

Rearrangement formulae

This section is not interesting and should not be read. Here we will prove formulas is which right hand side uses the same factors as the left hand side, just in different order. These facts are obvious in informal math sense, but Isabelle prover is not able to derive them automatically, so we have to prove them by hand.

Proving the facts about associative and commutative operations is quite tedious in formalized mathematics. To a human the thing is simple: we can arrange the elements in any order and put parantheses wherever we want, it is all the same. However, formalizing this statement would be rather difficult (I think). The next lemma attempts a quasi-algorithmic approach to this type of problem. To prove that two expressions are equal, we first strip one from parantheses, then rearrange the elements in proper order, then put the parantheses where we want them to be. The algorithm for rearrangement is easy to describe: we keep putting the first element (from the right) that is in the wrong place at the left-most position until we get the proper arrangement. As far removing parantheses is concerned Isabelle does its job automatically.

lemma (in group0 ) group0_4_L2:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G, d\in G, E\in G, F\in G shows (a\cdot b)\cdot (c\cdot d)\cdot (E\cdot F) = (a\cdot (d\cdot F))\cdot (b\cdot (c\cdot E))proof

Another useful rearrangement.

lemma (in group0 ) group0_4_L3:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G and A3: c\in G, d\in G, E\in G, F\in G shows a\cdot b\cdot ((c\cdot d)^{-1}\cdot (E\cdot F)^{-1}) = (a\cdot (E\cdot c)^{-1})\cdot (b\cdot (F\cdot d)^{-1})proof

Some useful rearrangements for two elements of a group.

lemma (in group0 ) group0_4_L4:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G shows b^{-1}\cdot a^{-1} = a^{-1}\cdot b^{-1}, (a\cdot b)^{-1} = a^{-1}\cdot b^{-1}, (a\cdot b^{-1})^{-1} = a^{-1}\cdot bproof

Another bunch of useful rearrangements with three elements.

lemma (in group0 ) group0_4_L4A:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G shows a\cdot b\cdot c = c\cdot a\cdot b, a^{-1}\cdot (b^{-1}\cdot c^{-1})^{-1} = (a\cdot (b\cdot c)^{-1})^{-1}, a\cdot (b\cdot c)^{-1} = a\cdot b^{-1}\cdot c^{-1}, a\cdot (b\cdot c^{-1})^{-1} = a\cdot b^{-1}\cdot c, a\cdot b^{-1}\cdot c^{-1} = a\cdot c^{-1}\cdot b^{-1}proof

Another useful rearrangement.

lemma (in group0 ) group0_4_L4B:
assumes P \text{ is commutative on } G and a\in G, b\in G, c\in G shows a\cdot b^{-1}\cdot (b\cdot c^{-1}) = a\cdot c^{-1} using prems , inverse_in_group , group_op_closed , group0_4_L4 , group_oper_assoc , inv_cancel_two

A couple of permutations of order for three alements.

lemma (in group0 ) group0_4_L4C:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G shows a\cdot b\cdot c = c\cdot a\cdot b, a\cdot b\cdot c = a\cdot (c\cdot b), a\cdot b\cdot c = c\cdot (a\cdot b), a\cdot b\cdot c = c\cdot b\cdot aproof

Some rearangement with three elements and inverse.

lemma (in group0 ) group0_4_L4D:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G shows a^{-1}\cdot b^{-1}\cdot c = c\cdot a^{-1}\cdot b^{-1}, b^{-1}\cdot a^{-1}\cdot c = c\cdot a^{-1}\cdot b^{-1}, (a^{-1}\cdot b\cdot c)^{-1} = a\cdot b^{-1}\cdot c^{-1}proof

Another rearrangement lemma with three elements and equation.

lemma (in group0 ) group0_4_L5:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G and A3: c = a\cdot b^{-1} shows a = b\cdot cproof

In abelian groups we can cancel an element with its inverse even if separated by another element.

lemma (in group0 ) group0_4_L6A:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G shows a\cdot b\cdot a^{-1} = b, a^{-1}\cdot b\cdot a = b, a^{-1}\cdot (b\cdot a) = b, a\cdot (b\cdot a^{-1}) = bproof

Another lemma about cancelling with two elements.

lemma (in group0 ) group0_4_L6AA:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G shows a\cdot b^{-1}\cdot a^{-1} = b^{-1} using prems , inverse_in_group , group0_4_L6A

Another lemma about cancelling with two elements.

lemma (in group0 ) group0_4_L6AB:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G shows a\cdot (a\cdot b)^{-1} = b^{-1}, a\cdot (b\cdot a^{-1}) = bproof

Another lemma about cancelling with two elements.

lemma (in group0 ) group0_4_L6AC:
assumes P \text{ is commutative on } G and a\in G, b\in G shows a\cdot (a\cdot b^{-1})^{-1} = b using prems , inverse_in_group , group0_4_L6AB , group_inv_of_inv

In abelian groups we can cancel an element with its inverse even if separated by two other elements.

lemma (in group0 ) group0_4_L6B:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G shows a\cdot b\cdot c\cdot a^{-1} = b\cdot c, a^{-1}\cdot b\cdot c\cdot a = b\cdot cproof

In abelian groups we can cancel an element with its inverse even if separated by three other elements.

lemma (in group0 ) group0_4_L6C:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G, d\in G shows a\cdot b\cdot c\cdot d\cdot a^{-1} = b\cdot c\cdot dproof

Another couple of useful rearrangements of three elements and cancelling.

lemma (in group0 ) group0_4_L6D:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G shows a\cdot b^{-1}\cdot (a\cdot c^{-1})^{-1} = c\cdot b^{-1}, (a\cdot c)^{-1}\cdot (b\cdot c) = a^{-1}\cdot b, a\cdot (b\cdot (c\cdot a^{-1}\cdot b^{-1})) = c, a\cdot b\cdot c^{-1}\cdot (c\cdot a^{-1}) = bproof

Another useful rearrangement of three elements and cancelling.

lemma (in group0 ) group0_4_L6E:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G shows a\cdot b\cdot (a\cdot c)^{-1} = b\cdot c^{-1}proof

A rearrangement with two elements and canceelling, special case of group0_4_L6D when c=b^{-1}.

lemma (in group0 ) group0_4_L6F:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G shows a\cdot b^{-1}\cdot (a\cdot b)^{-1} = b^{-1}\cdot b^{-1}proof

Some other rearrangements with four elements. The algorithm for proof as in group0_4_L2 works very well here.

lemma (in group0 ) rearr_ab_gr_4_elemA:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G, d\in G shows a\cdot b\cdot c\cdot d = a\cdot d\cdot b\cdot c, a\cdot b\cdot c\cdot d = a\cdot c\cdot (b\cdot d)proof

Some rearrangements with four elements and inverse that are applications of rearr_ab_gr_4_elem

lemma (in group0 ) rearr_ab_gr_4_elemB:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G, d\in G shows a\cdot b^{-1}\cdot c^{-1}\cdot d^{-1} = a\cdot d^{-1}\cdot b^{-1}\cdot c^{-1}, a\cdot b\cdot c\cdot d^{-1} = a\cdot d^{-1}\cdot b\cdot c, a\cdot b\cdot c^{-1}\cdot d^{-1} = a\cdot c^{-1}\cdot (b\cdot d^{-1})proof

Some rearrangement lemmas with four elements.

lemma (in group0 ) group0_4_L7:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G, d\in G shows a\cdot b\cdot c\cdot d^{-1} = a\cdot d^{-1}\cdot b\cdot c, a\cdot d\cdot (b\cdot d\cdot (c\cdot d))^{-1} = a\cdot (b\cdot c)^{-1}\cdot d^{-1}, a\cdot (b\cdot c)\cdot d = a\cdot b\cdot d\cdot cproof

Some other rearrangements with four elements.

lemma (in group0 ) group0_4_L8:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G, d\in G shows a\cdot (b\cdot c)^{-1} = (a\cdot d^{-1}\cdot c^{-1})\cdot (d\cdot b^{-1}), a\cdot b\cdot (c\cdot d) = c\cdot a\cdot (b\cdot d), a\cdot b\cdot (c\cdot d) = a\cdot c\cdot (b\cdot d), a\cdot (b\cdot c^{-1})\cdot d = a\cdot b\cdot d\cdot c^{-1}, (a\cdot b)\cdot (c\cdot d)^{-1}\cdot (b\cdot d^{-1})^{-1} = a\cdot c^{-1}proof

Some other rearrangements with four elements.

lemma (in group0 ) group0_4_L8A:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G, d\in G shows a\cdot b^{-1}\cdot (c\cdot d^{-1}) = a\cdot c\cdot (b^{-1}\cdot d^{-1}), a\cdot b^{-1}\cdot (c\cdot d^{-1}) = a\cdot c\cdot b^{-1}\cdot d^{-1}proof

Some rearrangements with an equation.

lemma (in group0 ) group0_4_L9:
assumes A1: P \text{ is commutative on } G and A2: a\in G, b\in G, c\in G, d\in G and A3: a = b\cdot c^{-1}\cdot d^{-1} shows d = b\cdot a^{-1}\cdot c^{-1}, d = a^{-1}\cdot b\cdot c^{-1}, b = a\cdot d\cdot cproof

end

Comments on AbelianGroup_ZF

Ring_ZF

Tue, 21 Oct 2008 00:13:00 GMT

theory Ring_ZF imports AbelianGroup_ZF

begin

This theory file covers basic facts about rings.

Definition and basic properties

In this section we define what is a ring and list the basic properties of rings.

We say that three sets (R,A,M) form a ring if (R,A) is an abelian group, (R,M) is a monoid and A is distributive with respect to M on R. A represents the additive operation on R. As such it is a subset of (R\times R)\times R (recall that in ZF set theory functions are sets). Similarly M represents the multiplicative operation on R and is also a subset of (R\times R)\times R. We don't require the multiplicative operation to be commutative in the definition of a ring.

Definition
IsAring(R,A,M) \equiv \text{IsAgroup}(R,A) \wedge (A \text{ is commutative on } R) \wedge
\text{IsAmonoid}(R,M) \wedge IsDistributive(R,A,M)

We also define the notion of having no zero divisors. In standard notation the ring has no zero divisors if for all a,b \in R we have a\cdot b = 0 implies a = 0 or b = 0.

Definition
HasNoZeroDivs(R,A,M) \equiv (\forall a\in R.\ \forall b\in R.\
M\langle a,b\rangle = TheNeutralElement(R,A) \longrightarrow
a = TheNeutralElement(R,A) \vee b = TheNeutralElement(R,A))

Next we define a locale that will be used when considering rings.

Locale ring0
assumes ringAssum: IsAring(R,A,M)
defines a + b \equiv A\langle a,b\rangle
defines ( - a) \equiv GroupInv(R,A)(a)
defines a - b \equiv a + ( - b)
defines a\cdot b \equiv M\langle a,b\rangle
defines 0 \equiv TheNeutralElement(R,A)
defines 1 \equiv TheNeutralElement(R,M)
defines 2 \equiv 1 + 1
defines a^2 \equiv a\cdot a

In the ring0 context we can use theorems proven in some other contexts.

lemma (in ring0 ) Ring_ZF_1_L1:
shows monoid0(R,M), group0(R,A), A \text{ is commutative on } R using ringAssum , IsAring_def , group0_def , monoid0_def

The additive operation in a ring is distributive with respect to the multiplicative operation.

lemma (in ring0 ) ring_oper_distr:
assumes A1: a\in R, b\in R, c\in R shows a\cdot (b + c) = a\cdot b + a\cdot c, (b + c)\cdot a = b\cdot a + c\cdot a using ringAssum , assms , IsAring_def , IsDistributive_def

Zero and one of the ring are elements of the ring. The negative of zero is zero.

lemma (in ring0 ) Ring_ZF_1_L2:
shows 0 \in R, 1 \in R, ( - 0 ) = 0 using Ring_ZF_1_L1 , group0_2_L2 , unit_is_neutral , group_inv_of_one

The next lemma lists some properties of a ring that require one element of a ring.

lemma (in ring0 ) Ring_ZF_1_L3:
assumes a\in R shows ( - a) \in R, ( - ( - a)) = a, a + 0 = a, 0 + a = a, a\cdot 1 = a, 1 \cdot a = a, a - a = 0 , a - 0 = a, 2 \cdot a = a + a, ( - a) + a = 0 using assms , Ring_ZF_1_L1 , inverse_in_group , group_inv_of_inv , group0_2_L6 , group0_2_L2 , unit_is_neutral , Ring_ZF_1_L2 , ring_oper_distr

Properties that require two elements of a ring.

lemma (in ring0 ) Ring_ZF_1_L4:
assumes A1: a\in R, b\in R shows a + b \in R, a - b \in R, a\cdot b \in R, a + b = b + a using ringAssum , assms , Ring_ZF_1_L1 , Ring_ZF_1_L3 , group0_2_L1 , group0_1_L1 , IsAring_def , IsCommutative_def

Cancellation of an element on both sides of equality. This is a property of groups, written in the (additive) notation we use for the additive operation in rings.

lemma (in ring0 ) ring_cancel_add:
assumes A1: a\in R, b\in R and A2: a + b = a shows b = 0 using assms , Ring_ZF_1_L1 , group0_2_L7

Any element of a ring multiplied by zero is zero.

lemma (in ring0 ) Ring_ZF_1_L6:
assumes A1: x\in R shows 0 \cdot x = 0 , x\cdot 0 = 0 proof

Negative can be pulled out of a product.

lemma (in ring0 ) Ring_ZF_1_L7:
assumes A1: a\in R, b\in R shows ( - a)\cdot b = - (a\cdot b), a\cdot ( - b) = - (a\cdot b), ( - a)\cdot b = a\cdot ( - b)proof

Minus times minus is plus.

lemma (in ring0 ) Ring_ZF_1_L7A:
assumes a\in R, b\in R shows ( - a)\cdot ( - b) = a\cdot b using assms , Ring_ZF_1_L3 , Ring_ZF_1_L7 , Ring_ZF_1_L4

Subtraction is distributive with respect to multiplication.

lemma (in ring0 ) Ring_ZF_1_L8:
assumes a\in R, b\in R, c\in R shows a\cdot (b - c) = a\cdot b - a\cdot c, (b - c)\cdot a = b\cdot a - c\cdot a using assms , Ring_ZF_1_L3 , ring_oper_distr , Ring_ZF_1_L7 , Ring_ZF_1_L4

Other basic properties involving two elements of a ring.

lemma (in ring0 ) Ring_ZF_1_L9:
assumes a\in R, b\in R shows ( - b) - a = ( - a) - b, ( - (a + b)) = ( - a) - b, ( - (a - b)) = (( - a) + b), a - ( - b) = a + b using assms , ringAssum , IsAring_def , Ring_ZF_1_L1 , group0_4_L4 , group_inv_of_inv

If the difference of two element is zero, then those elements are equal.

lemma (in ring0 ) Ring_ZF_1_L9A:
assumes A1: a\in R, b\in R and A2: a - b = 0 shows a=bproof

Other basic properties involving three elements of a ring.

lemma (in ring0 ) Ring_ZF_1_L10:
assumes a\in R, b\in R, c\in R shows a + (b + c) = a + b + c, a - (b + c) = a - b - c, a - (b - c) = a - b + c using assms , ringAssum , Ring_ZF_1_L1 , group_oper_assoc , IsAring_def , group0_4_L4A

Another property with three elements.

lemma (in ring0 ) Ring_ZF_1_L10A:
assumes A1: a\in R, b\in R, c\in R shows a + (b - c) = a + b - c using assms , Ring_ZF_1_L3 , Ring_ZF_1_L10

Associativity of addition and multiplication.

lemma (in ring0 ) Ring_ZF_1_L11:
assumes a\in R, b\in R, c\in R shows a + b + c = a + (b + c), a\cdot b\cdot c = a\cdot (b\cdot c) using assms , ringAssum , Ring_ZF_1_L1 , group_oper_assoc , IsAring_def , IsAmonoid_def , IsAssociative_def

An interpretation of what it means that a ring has no zero divisors.

lemma (in ring0 ) Ring_ZF_1_L12:
assumes HasNoZeroDivs(R,A,M) and a\in R, a\neq 0 , b\in R, b\neq 0 shows a\cdot b\neq 0 using assms , HasNoZeroDivs_def

In rings with no zero divisors we can cancel nonzero factors.

lemma (in ring0 ) Ring_ZF_1_L12A:
assumes A1: HasNoZeroDivs(R,A,M) and A2: a\in R, b\in R, c\in R and A3: a\cdot c = b\cdot c and A4: c\neq 0 shows a=bproof

In rings with no zero divisors if two elements are different, then after multiplying by a nonzero element they are still different.

lemma (in ring0 ) Ring_ZF_1_L12B:
assumes A1: HasNoZeroDivs(R,A,M), a\in R, b\in R, c\in R, a\neq b, c\neq 0 shows a\cdot c \neq b\cdot c using A1 , Ring_ZF_1_L12A

In rings with no zero divisors multiplying a nonzero element by a nonone element changes the value.

lemma (in ring0 ) Ring_ZF_1_L12C:
assumes A1: HasNoZeroDivs(R,A,M) and A2: a\in R, b\in R and A3: 0 \neq a, 1 \neq b shows a \neq a\cdot bproof

If a square is nonzero, then the element is nonzero.

lemma (in ring0 ) Ring_ZF_1_L13:
assumes a\in R and a^2 \neq 0 shows a\neq 0 using assms , Ring_ZF_1_L2 , Ring_ZF_1_L6

Square of an element and its opposite are the same.

lemma (in ring0 ) Ring_ZF_1_L14:
assumes a\in R shows ( - a)^2 = ((a)^2 ) using assms , Ring_ZF_1_L7A

Adding zero to a set that is closed under addition results in a set that is also closed under addition. This is a property of groups.

lemma (in ring0 ) Ring_ZF_1_L15:
assumes H \subseteq R and H \text{ is closed under } A shows (H \cup \{0 \}) \text{ is closed under } A using assms , Ring_ZF_1_L1 , group0_2_L17

Adding zero to a set that is closed under multiplication results in a set that is also closed under multiplication.

lemma (in ring0 ) Ring_ZF_1_L16:
assumes A1: H \subseteq R and A2: H \text{ is closed under } M shows (H \cup \{0 \}) \text{ is closed under } M using assms , Ring_ZF_1_L2 , Ring_ZF_1_L6 , IsOpClosed_def

The ring is trivial iff 0=1.

lemma (in ring0 ) Ring_ZF_1_L17:
shows R = \{0 \} \longleftrightarrow 0 =1 proof

The sets \{m\cdot x. x\in R\} and \{-m\cdot x. x\in R\} are the same.

lemma (in ring0 ) Ring_ZF_1_L18:
assumes A1: m\in R shows \{m\cdot x.\ x\in R\} = \{( - m)\cdot x.\ x\in R\}proof

Rearrangement lemmas

In happens quite often that we want to show a fact like (a+b)c+d = (ac+d-e)+(bc+e)in rings. This is trivial in romantic math and probably there is a way to make it trivial in formalized math. However, I don't know any other way than to tediously prove each such rearrangement when it is needed. This section collects facts of this type.

Rearrangements with two elements of a ring.

lemma (in ring0 ) Ring_ZF_2_L1:
assumes a\in R, b\in R shows a + b\cdot a = (b + 1 )\cdot a using assms , Ring_ZF_1_L2 , ring_oper_distr , Ring_ZF_1_L3 , Ring_ZF_1_L4

Rearrangements with two elements and cancelling.

lemma (in ring0 ) Ring_ZF_2_L1A:
assumes a\in R, b\in R shows a - b + b = a, a + b - a = b, ( - a) + b + a = b, ( - a) + (b + a) = b, a + (b - a) = b using assms , Ring_ZF_1_L1 , inv_cancel_two , group0_4_L6A

In commutative rings a-(b+1)c = (a-d-c)+(d-bc). For unknown reasons we have to use the raw set notation in the proof, otherwise all methods fail.

lemma (in ring0 ) Ring_ZF_2_L2:
assumes A1: a\in R, b\in R, c\in R, d\in R shows a - (b + 1 )\cdot c = (a - d - c) + (d - b\cdot c)proof

Rerrangement about adding linear functions.

lemma (in ring0 ) Ring_ZF_2_L3:
assumes A1: a\in R, b\in R, c\in R, d\in R, x\in R shows (a\cdot x + b) + (c\cdot x + d) = (a + c)\cdot x + (b + d)proof

Rearrangement with three elements

lemma (in ring0 ) Ring_ZF_2_L4:
assumes M \text{ is commutative on } R and a\in R, b\in R, c\in R shows a\cdot (b\cdot c) = a\cdot c\cdot b using assms , IsCommutative_def , Ring_ZF_1_L11

Some other rearrangements with three elements.

lemma (in ring0 ) ring_rearr_3_elemA:
assumes A1: M \text{ is commutative on } R and A2: a\in R, b\in R, c\in R shows a\cdot (a\cdot c) - b\cdot ( - b\cdot c) = (a\cdot a + b\cdot b)\cdot c, a\cdot ( - b\cdot c) + b\cdot (a\cdot c) = 0 proof

Some rearrangements with four elements. Properties of abelian groups.

lemma (in ring0 ) Ring_ZF_2_L5:
assumes a\in R, b\in R, c\in R, d\in R shows a - b - c - d = a - d - b - c, a + b + c - d = a - d + b + c, a + b - c - d = a - c + (b - d), a + b + c + d = a + c + (b + d) using assms , Ring_ZF_1_L1 , rearr_ab_gr_4_elemB , rearr_ab_gr_4_elemA

Two big rearranegements with six elements, useful for proving properties of complex addition and multiplication.

lemma (in ring0 ) Ring_ZF_2_L6:
assumes A1: a\in R, b\in R, c\in R, d\in R, e\in R, f\in R shows a\cdot (c\cdot e - d\cdot f) - b\cdot (c\cdot f + d\cdot e) =
(a\cdot c - b\cdot d)\cdot e - (a\cdot d + b\cdot c)\cdot f, a\cdot (c\cdot f + d\cdot e) + b\cdot (c\cdot e - d\cdot f) =
(a\cdot c - b\cdot d)\cdot f + (a\cdot d + b\cdot c)\cdot e, a\cdot (c + e) - b\cdot (d + f) = a\cdot c - b\cdot d + (a\cdot e - b\cdot f), a\cdot (d + f) + b\cdot (c + e) = a\cdot d + b\cdot c + (a\cdot f + b\cdot e)proof

end

Comments on Ring_ZF

Test Sliders

Thu, 16 Oct 2008 00:50:00 GMT

a \cdot b = b\cdot aproof
{

fix x assume x\in A
show x \in B

} thus A \subseteq B

StyleSheet

Mon, 29 Sep 2008 23:11:00 GMT

blockquote { margin-right:0; }