Logical deduction in predicate calculus from a reflexive predicate

91 Views Asked by At

I am trying to prove the following argument by means of deduction:

Laa ∴ ∃x∃yLxy

Naturally I understand the entailment - If there is a constant a for which the relation L occurs reflexively, there are some x,y (indeed x=y) between which the relation L occurs.

I cannot seem to understand how to travel from reflexivity to generality, because the next step only allows me to determine ∃xLxx.

If I try to assume the negative and prove by contradiction I still do not know how to proceed after ~(∃x∃yLxy).

Any ideas?

1

There are 1 best solutions below

1
On

The proof is simple, using two $\exists \text I$: from $\varphi[x/a]$, derive $\exists x \varphi$.

Thus, we have only to find the suitable $\varphi$; we have that $Laa := Lxy[x/a,y/a]$ and thus the derivation will be:

  1. $Laa$

  2. $\exists y Lay$ --- from 1) by $\exists \text I$

  3. $\exists x \exists y Lxy$ --- from 2) by $\exists \text I$.


If you cannot correctly manage the rule with some "proof assistant", you can prove it by contradiction, exploiting the equivalence between $\lnot \exists x \exists y Lxy$ and $\forall x \forall y \lnot Lxy$.

Instantiating the second one twice you get $\lnot Laa$ that contradicts the premise.