countermodel interpretation - another example with existential quantifiers

Question

Still struggling with interpretation in logic. I have this:

∀xp(x) ↔ ∀xq(x), ∀xq(x) ↔ ∀xr(x) |=∃x(p(x) ↔ r(x))

1. I first tried to solve it with a resolution method, and got stuck on this:

$ \lnot px \lor qx, px \lor \lnot qx $ from the first premise

$ rx \lor \lnot qx, qx \lor \lnot rx $ from the second

$ \lnot px \lor \lnot rx, px \lor rx $ from the negated conclusion

I could not find any atomic sentence to "eliminate" and don't know how to proceed.

2. I tried a "human interpretation", to search for values that would evaluate the premises as True and the conclusion as False.

I read it as:

• for all value x of P, and for all values x of Q they are equivalent when their values are the same (first premise)
• for all value x of Q, and for all values x of R they are equivalent when their values are the same (second premise)
• there exist one value that makes P equal to R

And I thought the above is true, since if P is equivalent to Q and Q to R, there must be some P equivalent to R.

But it is wrong because the solution shows a countermodel: Universe = {a,b} p: { b} r: { a } q: { }

As a test, I tried to change the conclusion to ∀x(p(x) ↔ r(x)) (and the countermodel is p: { 1 } r: { 0 } q: { 1 }

and to :

∃xp(x) ↔ ∃xr(x) and the countermodel changes to: p: { } r: { 0 } q: { }

while I was expecting at least being equal to ∃x( p(x) ↔ r(x) ).

• Can you show how to solve this step by step ?
• Can you help understand why the counter model changes in the examples above ?

Answer 1

$$∀xp(x) ↔ ∀xq(x),\:\: ∀xq(x) ↔ ∀xr(x) \:\models\: ∃x(p(x) ↔ r(x))$$

I read it as:

• for all value x of P, and for all values x of Q they are equivalent when their values are the same (first premise)

Try formalising your above translation. Do you see that the only sensible way to do so is: $$∀x \,\big(Px ↔ Qx\big)\,?$$ Instead, the first premise actually says:

• P and Q are both universally true or both not universally true.

Note that not being universally true is not logically equivalent to being universally false: $$¬∀x\,Px \quad\not\equiv\quad ∀x\,¬Px.$$

the solution shows a countermodel: $$\text{Universe}= \{a,b\}\\ p: \{ b\}\\ r: \{ a \}\\ q: \{ \}$$

I tried to change the conclusion to $∀x(p(x) ↔ r(x)),$ and the countermodel is $$p: \{ 1 \}\\r: \{ 0 \}\\q: \{ 1 \}$$

and to $∃xp(x) ↔ ∃xr(x),$ and the countermodel changes to $$p: \{ \}\\r: \{ 0 \}\\q: \{ \}$$ while I was expecting at least being equal to $∃x( p(x) ↔ r(x) ).$

Can you help understand why the counter model changes in the examples above ?

For all three countermodels, $P,Q,R$ are not universally true, so the two premises are satisfied.

1. The original conclusion wants some value that gives $P$ and $R$ the same truth value; the given countermodel indeed fails to satisfy that.
2. The second conclusion wants $P$ and $R$ to be universally equivalent; the given countermodel indeed fails to satisfy that.
3. The third conclusion wants $P$ and $R$ to each to be true for some value (not necessarily a shared value!) or each be universally false; the given countermodel indeed fails to satisfy that.

The following rewrites are helpful; here they clarify that the two copies of $x$ aren't necessarily the same object: $$∀x \,Px ↔ ∀x \,Qx \quad\equiv\quad ∀x \,Px ↔ ∀y \,Qy\\ ∃x \,Px ↔ ∃x \,Qx \quad\equiv\quad ∃x \,Px ↔ ∃y \,Qy$$