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Express this system specifications using predicates, quantifiers, and logical connectives, if necessary.

There are exactly two systems that monitor every remote server.

My solution is :

∃x∃y∀z(x != y ∧ (∀s M(z,s) → (z = x ∨ z = y))), where M(a, b) means that system a monitors remote server b

So can anyone please check my solution if it's correct and if not what is the correct solution?

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2 Answers 2

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The problem:

There are exactly two systems that monitor every remote server.

M(a, b) means that system a monitors remote server b.

Can be rewritten into:

$\color{red}{\text{There is a system x and there is a system z such that the two systems are different}}$ and $\color{blue}{\text{system x monitors all remote servers and system z monitors all remote servers}}$ and $\color{green}{\text{no other system w monitors all remote servers}}$.

$$\color{red}{\exists x,z(x\neq z} \land \color{blue}{\forall y M(x,y) \land \forall y M(z,y)}\land \color{green}{\lnot\exists w(w \neq x \land w \neq z \land \forall y M(w,y))}\color{red}{)}$$

Since $\lnot\exists w(P(w)) \iff \forall w(\lnot P(w))$, we can rewrite this as:

$$\color{red}{\exists x,z(x\neq z} \land \color{blue}{\forall y M(x,y) \land \forall y M(z,y)}\land \color{green}{\forall w(\underbrace{w = x \lor w = z}_{\psi} \lor \lnot\underbrace{\forall y M(w,y)}_{\phi})}\color{red}{)}$$

Finally, using the property:

$\psi \lor \lnot\phi \iff \phi\rightarrow \psi$

we end up with the accepted solution:

$$\color{red}{\exists x,z(x\neq z} \land \color{blue}{\forall y M(x,y) \land \forall y M(z,y)}\land \color{green}{\forall w(\forall y M(w,y) \rightarrow (w = x \lor w = z))}\color{red}{)}$$

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The valid solution is:

∃x∃y((x ≠ y) ∧ ∀zM(x,z) ∧ ∀zM(y,z) ∧ ∀w(∀zM(w,z) → (w = x) ∨ (w = y)))

And so my solution is too close to the right I just missed to put ∀zM(x,z) ∧ ∀zM(y,z)

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    $\begingroup$ In case you still want a formal proof for this, I think I finally got it. Also, thanks for letting me know when I was wrong. I have learned a couple of important things I didn't know before. $\endgroup$
    – mateleco
    Commented Apr 1, 2020 at 21:32
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    $\begingroup$ Thanks alot for your efforts to help me .. really happy for that discussion dear @mateleco $\endgroup$
    – Omar_Hafez
    Commented Apr 1, 2020 at 23:21

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