Distinct variables for different clauses

Question

In resolution theorem proving, it is normally assumed variables in different clauses are distinct. This is not something that happens automatically; it requires significant extra code and computation to implement. Given that, I'm looking for a test case for it.

The problem is, in all the test cases I've tried so far, it makes no difference. Presumably it matters only in unusual edge cases. As Wikipedia puts it, "variables in different clauses are distinct... Now, unifying Q(X) in the first clause with Q(Y) in the second clause means that X and Y become the same variable anyway."

Are there any known test cases that will actually give the wrong answer if different clauses use the same variables?

Answer 1

Edit: I found a better example. Consider these clauses: \begin{align} & \lnot P(x) \lor P(f(x)) \\ & P(x) \\ & \lnot P(f(f(x)))\\ \end{align} Obviously, this set of clauses is contradictory. But without renaming variables, the only possible resolvent is $P(f(x))$ and no more resolvents are possible - all lead to substituting $f(x)$ for $x$, which is impossible.

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Edit: Consider the meaning of clauses. Each clause is implicitly universally quantified. So the meaning of its variables isn't fixed to anything. Now let's say you have two clauses both containing $x$. If you perform resolution without renaming $x$ in one of them, then you add a meaning to $x$ which it doesn't have: you say that $x$ means the same thing in both clauses, which is not true. If you don't have distinct variables in your clauses, resolution will give you too weak conclusions.

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(The original answer.) For example, let's have 4 clauses:

1. $A\lor B(x)$
2. $\lnot A\lor C(x)$
3. $\lnot B(c)$
4. $\lnot C(d)$

where $x, y$ are variables and $c, d$ constants. If we perform resolution on the first two without renaming $x$, we'll get $B(x)\lor C(x)$. We can proceed with $\lnot B(c)$ to get $C(c)$ but now we cannot resolve it with $\lnot C(d)$.

On the other hand, if we rename $x$ to $y$ in the second one to have disjoint set of variables, we'll get $B(x)\lor C(y)$ from the first resolution step and we can derive an empty clause using $\lnot B(c)$ and $\lnot B(d)$.