Nested Derivations

Sometimes, a conditional derivation will not be enough to show that a certain conditional we want is true. The trouble is that in a conditional derivation, you can only make one assumption.

But there are situations where it seems that we ought to make more than one assumption. For example, conditionals of the form P → (Q → R) will be difficult to prove by assuming only P. Such a conditional says that if P is true, then if, furthermore, R is true, then Q is true. Intuitively, this seems to say that Q is true whenever P and Q are both true. So, to prove that conditional, it should be enough to demonstrate that Q is true in any hypothetical situation where R and Q are true. But this means assuming two things: both R and Q.

This need for multiple simultaneous assumptions is reflected in ordinary informal reasoning.

We may, while playing chess, want to know whether a statement like “If I take the queen, then if my opponent moves her bishop to 8D, she can put me in check” is true. Just imagining that we take the queen is not enough for us to figure out what will happen if we take the queen and our opponent moves her bishop. We need to first imagine that we take the queen. We then need to add another piece of information to that hypothetical scenario—we need suppose additionally that our opponent then moves her Bishop.

What we are doing is adding another “layer” of hypothesis to our imaginary situation. Just as our first imaginary situation was like the real world except for one addition (that we took the queen), our second imaginary situation is just like our first imaginary situation, but with one addition (that our opponent moves her bishop).

It is natural to model this type of “double assumption”, by allowing for derivations to occur inside of other derivations. If you like a good metaphor, you can think of a double assumption as a “dream within a dream.” We first imagine that we are in a scenario where P is known to be true. We then imagine that, in this hypothetical scenario, we are imagining that Q is known to be true. If our hypothetical reasoning within the hypothetical scenario where P is true lets us deduce that Q → R is true in that scenario, then we can (in real life) infer that P → (Q → R) is true. Let’s call a derivation that has other derivations occurring inside of it a Nested Derivation. And let’s divide the nested derivations we know how to construct into two types: Nested Conditional Derivations, and Nested Direct Derivations.

Available Lines

The main difficulty with this kind of “doubly hypothetical reasoning” is keeping track of what you are currently allowing yourself to assume.

The difficulty is that everything that is shown given those two hypotheses is only true if those two hypotheses are true. And we cannot use this “conditional” information to draw unqualified conclusions, any more than we can use facts that are true only within our daydreams to plan our morning commute or our finances.

For example, compare the following two pseudo-arguments:

1. Show: If santa is real, then if I am good I will get a pony.
2.    Santa is real                                    :AS
3.    Show: If I am good, I will get a pony 
4.        I am good                                    :AS
5.        if Santa is real, good children are rewarded :PR
6.        good children are rewarded                   :MP 2 5
7.        I will get a pony                            :? 4 6
8.    :CD 7
9. :CD 3

This one seems good, if we’re willing to grant the premise, and the mysterious inference on line 7 (this is an example of an inference involving quantifiers—something we will cover later in this course).

1. Show: If Santa is real, then I will get a pony.
2.    Santa is real                                    :AS
3.    Show: If I am good, I will get a pony 
4.        I am good                                    :AS
5.        if Santa is real, good children are rewarded :PR
6.        good children are rewarded                   :MP 2 5
7.        I will get a pony                            :? 4 6
8.    :CD 7
9. :CD 7

This argument seems questionable. Santa, even if he exists, only rewards good children. So, even if Santa exists, that doesn’t guarantee that I will get a pony—I might have been bad, in which case, I’ll get nothing, or coal.

The trouble is with line 9. To justify finishing our proof, we cite line 7. But line 7 is only true within the hypothetical scenario where we suppose that I am good—it stops being true when we leave that scenario and move to the scenario where all that is known is that Santa is real. So it can’t be used to draw a conclusion in that outer scenario.

In order to avoid mistakes like this one, we need some way of thinking about which lines can be cited, and which lines cannot be cited in the course of a proof. It seems that the ones that we cannot cite are the ones that are parts of hypothetical scenarios that we have already stopped considering—those scenarios whose truths are contained between a show line and a corresponding QED line, like CD or DD. So, let us use this criterion to distinguish between the available lines and the unavailable lines.

Unavailable Line

A line is unavailable at a certain point in a proof if it is contained between a show line and QED line corresponding to that show line which are both earlier in the proof.

Having said what an unavailable line is, we can now say what we will officially count as a legitimate nested derivation.

Nested Direct Derivation

A Nested Direct Derivation is is a sequence of assertions, aimed at showing some statement φ, each of which is justified, either because

  • it is a premise of the argument under consideration;

  • it is the conclusion of a rule of direct derivation applied to previous assertions which are available from the line it occupies;

  • or, it is listed next to a show-line that contains a complete derivation and is justified by a matching QED line.

We use nested direct derivations in the same way that we use ordinary direct derivations. We begin with a show line, saying what it is we aim to show, for example, φ. We then indent, and beneath our indentation, we place our nested direct derivation. Once we arrive at φ, within the derivation, we can finish by pointing to that line from a qed line containing the justification DD, and the line number on which φ occurs. The only complication when the proof involves nesting is that we must also be sure that φ is available from qed line.

Nested Conditional Derivation

A Nested Conditional Derivation is is a sequence of assertions aimed at showing some statement φ → ψ, beginning with an assumption that φ, in which every assertion after the first is justified, either because

  • it is a premise of the argument under consideration;

  • it is the conclusion of a rule of direct derivation applied to previous assertions;

  • or, it is listed next to a show-line that contains a complete derivation and is justified by a matching QED line.

As with nested direct derivations, we can use nested conditional derivations in almost the same way as their unnested counterparts. The main difference is that we must be sure, when we write down our finishing QED line, that the line we cite is available from the QED line.

So, to recap, the main difference between the nested and the simple cases is that in the nested case, we allow a new way of basing later assertions on earlier assertions. In addition to rules of direct inference, we allow for an assertion to be made on the basis of a earlier assertions if that assertion can be shown, by a derivation which uses those earlier assertions.

Allowing these new assertions requires some new safety measures, to make sure that we don’t mix up what we known only conditionally with what we really know. We must make sure that we make assertions only on the basis of lines that are available from the point where we wish to make the assertion, and close derivations only on the basis of lines that are available from where we want to close the derivation.

Here are some examples of nested derivations.

For the argument P → (Q → R), ¬Q → S ⊢ P → (¬S → R), we can derive:

1. Show: P-> (~S-> R)
2.    P              :AS
3.    P -> (Q -> R)    :PR
4.    ~Q -> S         :PR
5.    Q -> R          :MP 2 3
6.    Show: ~S -> R          
7.      ~S           :AS
8.      ~~Q          :MT 7 4
9.      Q            :DN 8
10.     R            :MP 9 5
11.   :CD 10
12. :CD 6

For the argument (P → R)→S . P → Q . Q → R∴ S, we can derive:

1. Show: S
2.   (P -> R)-> S    :PR
3.   P -> Q          :PR
4.   Q -> R          :PR
5.   Show: P -> R   
6.       P           :AS
7.       Q           :MP 3 6
8.       R           :MP 4 7
9.   :CD 8
10.  S               :MP 2,5
11.:DD 10

Problem Set 6

Construct derivations to show the validity of the listed arguments.

Abbreviations are the same as in previous chapters. When the argument turns a light green color, then you can press the “submit” button to submit your work.

exercise 6.1
6.1 :|-: P->(Q->P)
exercise 6.2
6.2 P->Q :|-: (Q->R)->(P->R)
exercise 6.3
6.3 P->(Q->R) :|-: (P->Q) -> (P->R)
exercise 6.4
6.4 P->(Q->R) :|-: Q->(P->R)
exercise 6.5
6.5 (S->R)->Q, R :|-: Q

Please also complete the translations below, using the following scheme of abbreviation

W  =  William leaves
V  =  Veronica leaves
U  =  Ursula leaves

exercise 6.6
[65,79,90,76,73,36,90,95,65,58,73]
exercise 6.7
[65,79,91,76,73,37,87,76,82,76,65,94,32,65,65,82,73,38,94,65]
exercise 6.8
[65,79,84,76,73,91,90,54,76,65,87,83,90,55,69,76,68,77,87,76,57,76]
exercise 6.9
[65,79,85,76,73,91,90,55,76,65,87,83,90,54,69,76]
exercise 6.10
[65,79,93,92,73,36,90,95,76,68,63,94,73,52,69,76]