Logical Consequences and Ordered Fields.

Question

How do I show that these two:

$1.$ $\forall x(0 < x \rightarrow (-x) < 0)$

$2.$ $\forall x \forall y \forall z((x<y \wedge z<0) \rightarrow (y *z) <(x*z))$

are logical consequences of the theory of ordered fields?

Like, I can see why it's a consequence... I just don't know how to prove that it's a logical consequence. Any help?

Answer 1

In Derek Goldrei, Propositional and Predicate Calculus : A Model of Argument (2005)we have :

• first-order axioms for commutative rings : Ax.1 to Ax.8 [page 202]

• additional axioms for fields : Ax.10 and Ax.11 [page 203]

• additional axioms for ordered fields : Ax.12 to Ax.16 [page 205].

For

$1. \quad ∀x (0 < x → (−x) < 0)$.

From Ax.15 : $∀x∀y∀z(x < y → (x + z) < (y + z))$

we have :

$∀y∀z(0 < y → (0 + z) < (y + z))$

and so :

$0 < x → (0 + (-x)) < (x + (-x))$.

From Ax.3 : $∀x((x + (−x)) = 0 ∧ ((−x) + x) = 0)$

we have : $(x + (−x)) = 0$;

thus, substituting into previous formula :

$0 < x → (0 + (-x)) < 0$.

Form Ax.2 : $∀x((x + 0) = x ∧ (0 + x) = x)$

we have : $0 + (-x) = (-x)$.

Substituting again, we have :

$0 < x → (-x) < 0$

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For

$2. \quad ∀x∀y∀z((x<y∧z<0)→(y∗z)<(x∗z))$

consider $x<y$ and $z<0$ and assume that : $\lnot ((y∗z)<(x∗z))$.

From Ax.14 : $∀x∀y(x < y ∨ (x = y ∨ y < x))$ we have that : either $(y∗z)=(x∗z)$ or $(x∗z)<(y∗z)$.

Assume (i) : $(y∗z)=(x∗z)$; by the previous proved fact [see Exercise 4.90 (b), page 204] that : $∀x∀y∀z((x \ne 0) \rightarrow ((x ∗ y) = (x ∗ z) → y = z))$ and Ax.8 : $∀x∀y (x ∗ y) = (y ∗ x)$, from the fact that $z \ne 0$ [because $z < 0$] we conclude with : $x=y$.

Thus, by substitution into assumption $x<y$, we get : $x < x$, contrary to Ax.12 : $∀x ¬(x < x)$.

Assume (ii) : $(x∗z)<(y∗z)$. In the same way used to prove : $∀x (0 < x → (−x) < 0)$ we can prove : $∀x (x < 0 → 0 < (−x))$.

Thus, from assumption $z < 0$ we have $0 < (-z)$.

Using Ax.16 : $∀x∀y∀z((x < y ∧ 0 < z) → (x ∗ z) < (y ∗ z))$ we get :

$(x ∗ (-z)) < (y ∗ (-z))$.

We have assumed : $(x∗z)<(y∗z)$; thus $(x∗z) + (x ∗ (-z)) < (y∗z) + (x ∗ (-z))$ from Ax.15 : $∀x∀y∀z(x < y → (x + z) < (y + z))$.

By distributive law : Ax.6 : $∀x∀y∀z (x ∗ (y + z)) = ((x ∗ y) + (x ∗ z))$ and Ax.3 : $∀x((x + (−x)) = 0$, we have :

$(x∗z) + (x ∗ (-z)) = (x ∗(z + (-z)) = (x ∗ 0) < (y∗z) + (x ∗ (-z))$.

With the previous proved fact [see Exercise 4.88 (a), page 203] that : $∀x (x ∗ 0) = 0$ we have :

$0 < (y∗z) + (x ∗ (-z))$.

In the same way, from $(x ∗ (-z)) < (y ∗ (-z))$ we get : $(x ∗ (-z)) + (y∗z) < (y ∗ (-z)) + (y∗z)$, by Ax.15, and again we use distributivity to have :

$(x ∗ (-z)) + (y∗z) < 0$.

Thus, from : $0 < (y∗z) + (x ∗ (-z))$ and $(x ∗ (-z)) + (y∗z) < 0$, with Ax.13 : $∀x∀y∀z((x < y ∧y < z) → x < z)$, we conclude with :

$0 < 0$

that contradicts Ax12 : $∀x ¬(x < x)$.

Resuming : with assumptions $x<y, z<0$ and $\lnot ((y∗z)<(x∗z))$ we have derived a contradiction.

Thus :

$x<y, z<0 \vDash ((y∗z)<(x∗z))$

i.e.

$(x<y \land z<0) \rightarrow ((y∗z)<(x∗z))$.