# 6.3: Logical Truth, Contradictions, Inconsistency, and Logical Equivalence

This section straightforwardly applies concepts you have already learned for sentence logic. We said that a sentence of sentence logic is a logical truth if and only if it is true in all cases, that is, if and only if it comes out true for all assignments of truth values to sentence letters. The concept of logical truth is the same in predicate logic if we take our cases to be interpretations of a sentence:

A closed predicate logic sentence is a Logical Truth if and only if it is true in all its interpretations.

Proof of logical truth also works just as it did for sentence logic, as we discussed in section 7-3 of Volume I. A derivation with no premises shows all its conclusions to be true in all cases (all assignments of truth values to sentence letters in sentence logic, all interpretations in predicate logic). A brief reminder of the reason: If we have a derivation with no premises we can always tack on unused premises at the beginning of the derivation. But any case which makes the premises of a derivation true makes all the derivation's conclusions true. For any case you like, tack on a premise in which that case is true. Then the derivation's conclusions will be true in that case also:

A derivation with no premises shows all its conclusions to be logical truths.

Contradictions in predicate logic also follow the same story as in sentence logic. The whole discussion is the same as for logical truth, except that we replace "true" with "false":

A closed predicate logic sentence is a Contradiction if and only if it is false in all its interpretations.

To demonstrate a sentence, X, to be a contradiction, demonstrate its negation, ~X, to be a logical truth. That is, construct a derivation with no premises, with ~X as the final conclusion.

If you did exercise 7-5 (in volume I), you learned an alternative test for contradictions, which also works in exactly the same way in predicate logic:

A derivation with a sentence, X, as its only premise and two sentences, Y and ~Y, as conclusions shows X to be a contradiction.

Exercise 7-8 (volume I) dealt with the concept of inconsistency. Once more, the idea carries directly over to predicate logic. I state it here, together with several related ideas which are important in more advanced work in logic:

A collection of closed predicate logic sentences is Consistent if there is at least one interpretation which makes all of them true. Such an interpretation is called a Model for the consistent collection of sentences. If there is no interpretation which makes all of the sentences in the collection true (if there is no model), the collection is Inconsistent.

A finite collection of sentences is inconsistent if and only if their conjunction' is a contradiction.

To demonstrate that a finite collection of sentences is inconsistent, demonstrate their conjunction to be a contradiction. Equivalently, provide a derivation with all of the sentences in the collection as premises and a contradiction as the final conclusion.

Finally, in predicate logic, the idea of logical equivalence of closed sentences works just as it did in sentence logic. We have already discussed this in section 3-4:

Two closed predicate logic sentences are Logically Equivalent if and only if in each of their interpretations the two sentences are either both true or both false.

Exercise 4-3 (volume I) provides the key to showing logical equivalence, as you already saw if you did exercise 7-9 (volume I). Two sentences are logically equivalent if in any interpretation in which the first is true the second is true, and in any interpretation in which the second is true the first is true. (Be sure you understand why this characterization comes to the same thing as the definition of logical equivalence I just gave.) Consequently

To demonstrate that two sentences, X and Y, are logically equivalent, show that the two arguments, "X. Therefore Y." and "Y. Therefore X." are both valid. That is, provide two derivations, one with X as premise and Y as final conclusion and one with Y as premise and X as final conclusion.

Exercise

6-3. Provide derivations which show that the following sentences are logical truths:

a) (Vx)(Vy)Lxy ⊃ (∃x)(∃y)Lxy
b) (Vx)(Gx v ~Gx)
c) (Vx)(∃y)(Ax & By) ⊃ (∃x)(Ax & Bx)
d) (∃y)[ky & (Vx)(Dx ⊃ Rxy)] ⊃ (Vx)[Dx ⊃ (∃y)(ky & Rxy)]
e) (∃x)(Vy)(Fy ⊃ Fx)
f) (Vx)(∃y)(Fy ⊃ Fx)
g) (∃x)(Vy)(Fx ⊃ Fy)

6-4. Provide derivations which show that the following sentences are contradictions:.

a) (Vx)(Ax ⊃ Bx) & (∃x)(~Bx & (Vy)Ay)
b) (Vx)(Rxb ⊃ ~Rxb) & (∃x)Rxb
c) (Vx)(Vy)Lxy & (∃y)~Lyx]
d) (Vx)(∃y)(Mx & ~My)
e) (Vx)(∃x)(Vw)(∃z)(Lxw & ~Lyz)

6-5. Provide derivations which show that the following collections of sentences are inconsistent:

a) (Vx)Kx, (Vy)~(ky v Lya)
b) (Vy)(∃y)Rxy, (∃x)(Vy)~Rxy
c) (∃x)Dx, (Vx)(Dx ⊃ (Vy)(Vz)Ryz), (∃x)(∃y)~Rxy
d) (∃x)(∃y)(Rxx & ~Ryy & Rxy), (Vx)(Vy)(Rxy ⊃ Ryx), (Vx)(Vy)(Vz)[(Rxy & Ryz) ⊃ Rxz)]

6-6. a) List the pairs of sentences which are shown to be logically equivalent by the examples in this chapter and any of the derivations in exercises 6-1 and 6-8.
b) Write derivations which show the following three arguments to be valid. (You will see in the next part of this exercise that there is a point to your doing these trivial derivations.)

(Vx)Rxa (Vx)Rxx (Vx)Px
(∃x)Rxa (∃x)Rxx (∃x)Px

c) Note that the three derivations you provided in your answer to (b) are essentially the same. From the point of view of these derivations, 'Rxa' and 'Rxx' are both open sentences which we could have just as well have written as P(u), an arbitrary (perhaps very complex) open sentence with u as its only free variable. In many of the problems in 5-5 and 5-7, I threw in names and repeated variables which played no real role in the problem, just as in the first two derivations in (b) above. (I did so to keep you on your toes in applying the new rules.) Find the problems which, when recast in the manner illustrated in (b) above, do the work of proving the following logical equivalences. Here, P(u) and Q(u) are arbitrary open sentences with u as their only free variable. A is an arbitrary closed sentence.

(Vu)(P(u) & Q(u)) is logically equivalent to (Vu)P(u) & (Vu)Q(u)
(∃u)(P(u) v Q(u)) is logically equivalent to (∃u)P(u) v (∃u)Q(u)
A ⊃ (Vu)P(u) is logically equivalent to (Vu)(A ⊃ P(u))
A ⊃ (∃u)P(u) is logically equivalent to (∃u)(A ⊃ P(u))
(Vu)P(u) ⊃ A is logically equivalent to (∃u)(P(u) ⊃ A)
(∃u)P(u) ⊃ A is logically equivalent to (Vu)(P(u) ⊃ A)

d) Prove, by providing a counterexample, that the following two pairs of sentences are not logically equivalent. (A counterexample is an interpretation in which one of the two sentences is true and the other is false.)

(Vx)(Px v Qx) is not logically equivalent to (Vx)Px v (Vx)Qx
(∃x)(Px & Qx) is not logically equivalent to (∃x)Px & (∃x)Qx

e) Complete the work done in 6-l(c) and (d) to show that the following pairs of sentences are logically equivalent. (R is an arbitrary open sentence with u and v as its only two free variables.)

(Vu)(Vv)R(u, v) is logically equivalent to (Vv)(Vu)R(u, v)
(∃u)(∃v)R(u, v) is logically equivalent to (∃v)(∃u)R(u, v)

6-7. Here are some harder arguments to prove valid by providing derivations. In some cases it is easier to find solutions by using the derived rules for negated quantifiers. But in every case you should look for elegant solutions which do not use these rules.

a) (Vx)[(∃y)(Lxy v Lyx) ⊃ Lxx]
(∃x)(∃y)Lxy
(∃x)Lxx
(Everyone who loves or is loved by someone loves themself. Someone loves someone. Therefore, someone loves themself.)

b) (Vx)(Hx ⊃ Ax)
(Vx)[(∃y)(Hy & Txy) ⊃ (∃y)(Ay & Txy)]
(Horse are animals. Therefore horses' tails are animal's tails)

c) (Vx)(Vy)[(∃z)Lyz ⊃ Lxy]
(∃x)(∃Y)Lxy
(Vx)(Vy)Lxy
(Everyone loves a lover. Someone loves someone. Therefore, everyone loves everyone.)

d) (Vx)(Vy)([(∃x)(Rxy & ~Rxz) ⊃ Lxy]
~(∃x)Lxx
(Vx)(Vy)(~Ryx ⊃ ~Rxy)

e) (Vx){(∃x)Lxy 3 (∃x)[(Vz)Lyz & Lxy]}
(∃x)(∃y)Lxy
(∃x)(Vy)Lxy
(Everyone who loves someone loves someone who loves everyone. Someone loves someone. Therefore, someone loves everyone.)

f) (Vx)[Px ⊃ (Vy)(Hy ⊃ Rxy)]
(∃x)(Px & (∃y)~Rxy)
~(Vx)Hx

g) (Vx)[(Ex ⊃ (Vy)(Hy ⊃ Wxy)]
(∃x)[Hx & (VY)(Dy ⊃ Wxy)]
(Vx)(Vy)(Vz)[(Wxy & Wyz) ⊃ Wxz
(Vx)[∃x ⊃ (Vy)(Dy ⊃ Wxy)]
(Any elephant weighs more than a horse. Some horse weighs more than any donkey. If a first thing weighs more than a second, and the second weighs more than a third, the first weighs more than the third. Therefore, any elephant weighs more than any donkey.)

h) (Vx)(∃y)(Py ⊃ Qx)
(∃y)(Vx)(Py ⊃ Qx)
Note that in general a sentence of the form (Vx)(∃y)X does not imply a sentence of the form (∃y)(Vx)X (See problem 6 -1(e)). However, in this case, the special form of the conditional makes the argument valid.

i) (∃x)Px ⊃ (∃x)Qx
(Vx)(∃y)(Px ⊃ Qy)

j) (∃y)(Px ⊃ (Vx)Qx
(Vx)(Vy)(Px ⊃ Qy)

k) (Vx){Bx ⊃ [(∃y)Lxy ⊃ (∃y)Lyz]}
(Vx)[(∃y)Lyx ⊃ Lxx]
~(∃x)Lxx
(Vx)(Bx ⊃ (Vy)~Lxy)
(All blond lovers are loved. All those who are loved love themselves. No one loves themself. Therefore, all blonds love no one.)

l) (Vx){Fx ⊃ [Hx & (~Cx & ~Kx)]}
(Vx)[(Hx & ~(∃y)Nxy) ⊃ Dx]
(Vx)(Fx ⊃ (∃y)Nxy)

m) (Vy)(Cy ⊃ Dy)
(Vx)(∃y)[(Hx & Cx) & (Gy & Ryx)]
(∃x)Dx ⊃ (Vy)(Vz)(Ryz ⊃ Dy)
(∃x)(Gx & Cx)

n) (Vx)(Vy){(Rdy & Rxd) ⊃ Rxy]
(Vx)(Bx ⊃ Rdx)
(∃x)(Bx & Rxd)
(∃x)[Bx & (Vy)(By ⊃ Rxy)]

chapter review Exercise

Write short explanations in your notebook for each of the following.