Are the following relations reflexive/irreflexive/neither? Symmetric/asymmetric/neither? Transitive/intransitive/neither?

1) x is a biological father of y

2) x is between point a and y. (Here, let the domain D = {x | x is a geometrical point in 3-dimensional space}. 'a' names some arbitrary fixed geometrical point. Also, let the between-ness relation mean that x lies on a straight line between a and y).

3) x formally implies y.

4) x and y are formally consistent.

5) x isn't identical to y.

  • What do you think? What progress have you made so far? – Hunan Rostomyan Apr 12 '14 at 18:23
  • Number one I have irreflexive , asymmetric and intransitive. – Achilles Apr 12 '14 at 18:30
  • 2) im confused with , 3) I believe is reflexive, symmetric, transitive. I stopped at this point because im unsure if I'm doing it correctly or not. – Achilles Apr 12 '14 at 18:32

We begin by recalling the basic definitions needed to settle the questions:


R is reflexive       =def   ∀w : wRw;
R is symmetric    =def   ∀w, v : (wRv → vRw);
R is transitive      =def   ∀w, v, u : (wRv ∧ vRu) → wRu.

Now we identify the relations in questions (1–5), ignoring the other details:

Question 1. R(x,y) ≡ BiologicalFatherOf(x, y).

You're right about R being irreflexive and asymmetric, because no person is a biological father of himself and no person is a biological father of his father. But R is also intransitive, because the biological father of your biological father is not your biological father!

Question 2. R(x, y) ≡ [x ∈ (a..y)], where a is fixed and (x..y) is the straight line from x to y inclusive.

Is R reflexive, i.e., is R(x,x) true? R(x,x) is true iff x ∈ (a..x), i.e., iff x is on the straight line from a to x inclusive (Sorry for the unorthodox notation; I hope its use is justified here). Because of the inclusiveness, x is indeed on that line, so R is reflexive.

To check for symmetry, assume R(x,y) and ask is R(y,x) true? R(x,y) implies that x ∈ (a..y), i.e., x is on the straight line from a to y. R(y,x) says that y ∈ (a..x), i.e., that y is on the straight line from a to x. R is not symmetric, because the line from a to y might be longer from the line from a to x.

To check for transitivity, assume R(x,y) and R(y,z) and ask is R(x,z) true? R(x,y) means that x is on the straight line between a and y, and R(y,z) means that y is on the straight line from a to z, so the line from a to z is an extension of the line from a to y. It's given that x is between a and y, and since the line from a to z extends the line from a to y, x must also be between a and z! So R is transitive.
† Thanks to Lucas for directing me toward the right interpretation of the relation.

Question 3. R(x, y) ≡ (x → y).

You're right about reflexivity and transitivity, but wrong about symmetry. R is reflexive because every statement x materially implies itself. R can be seen to be transitive by two applications of modus ponens or detachment. R is certainly not symmetric, otherwise material implication collapses to equivalence.

Question 4. R(x, y) ≡ Consistent(x, y).

R is reflexive, because any statement is consistent with itself (I'm not sure about contradictions; should we exclude them or say that two contradictions are somehow consistent?). R is symmetric, because if x is consistent with y, then y is consistent with x. R is not transitive, because while x may be consistent with y, x may be inconsistent with some z consistent with y. That is because for x and y to be consistent, (x ∧ y) has to be satisfiable, which may be true for different reasons for (x and y) and (y and z), so (x and z) might still turn out to be unsatisfiable.

Question 5. R(x, y) ≡ (x != y).

R is irreflexive, because every x is indeed identical to itself. R is symmetric because if x isn't identical to y, then y is also not identical to x. Lastly, R is not transitive, because while (x and y) and (y and z) might be distinct, it doesn't imply that (x and z) are distinct, i.e., (x = z) is possible.

  • You've misread the definition of 2, should be "x is between a and y" not "a is between x and y" – Lucas Apr 12 '14 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.