Working with Multiple Assumptions

        We are not restricted in a derivation to a single assumption. There can be as many as you want or need. In some cases the assumptions will not be embedded in one another. Here is one such case. We are showing that ~(P v Q) I- ~P & ~Q.

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~(P v Q)
     P
     P v Q
     (P v Q) & ~(P v Q)
~P
     Q
     P v Q
     (P v Q) & ~(P v Q)
~Q
~P & ~Q

Premise
* A
* 2 vI
* 1, 3 & I
2-4 ~I
* A
* 6 vI
* 1, 7 &I
6-8 ~I
5, 9 &I


Note that at line 5 and below lines 2 through 4 are inaccessible. The subderivation has been discharged. Those lines cannot be appealed to at any later point. Recall that in the builder inaccessible lines are marked off with an *. Here is another case. We are showing that P & Q I- P ‹-› Q. Generally we will get to biconditionals by obtaining the two conditionals. In this particular case we will apply &E twice as the initial steps.

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P & Q
P
Q
     P
     Q
P -› Q
     Q
     P
Q -› P
(P -› Q) & (Q -› P)
P ‹-› Q

Premise
&E
&E
* A
* 3 Reit
4-5 -›I
* A
* 2 Reit
7-8 -›I
6, 9 &I
10 ‹-›I,E


        Typically, though not invariably, we will obtain a conclusion that is a conjunction by obtaining each of the conjuncts and applying &I. Here is an example of such a case that involves making multiple assumptions. We will show that P -› ~P, ~(Q v ~R) I- ~P & R

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P -› ~P
~(Q v ~R)
     P
     ~P
     P & ~P
~P
     ~R
     Q v ~R
     (Q v ~R) & ~(Q v ~R)
R
~P & R

Premise
Premise
* A
* 1, 3 -›E
* 3, 4 &I
3-5 ~I
* A
* 7 vI
* 2, 8 &I
7-9 ~E
6, 10 &I


At this point you should:

Go to Lesson 1

        But assumptions may be embedded. That is, you may have assumptions within assumptions. Here is one example where we wish to show that P |- R -› (Q -› P):

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P
     R

Premise
A


Since the sentence that we wish to derive is a conditional we set up a conditional introduction. But the consequent of that conditional is itself a conditional so we set up another conditional introduction to obtain it.

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P
     R
          Q

Premise
A
A



Notice that since we have a new assumption we must indent further. This provides a visual means of keeping track of assumptions. Let us move on.

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P
     R
          Q
          P
     Q -› P

Premise
A
* A
* 1 Reit
3-4 -›I


Notice that at this point we have terminated one assumption. Note that the *'s indicate that at line 5 lines 3 and 4 have become inaccessible. It is important to keep in mind, when working with embedded assumptions, that you can only terminate the nearest unterminated assumption above. The following is, for example, a mistake:

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P
     R
          Q
          P
     R -› P

Premise
* A
* A
* 1 Reit
2-4 -›I





Erroneous


The assumption made at line 2 cannot be terminated until the assumption immediately below it has been terminated. Here is a correct completed derivation:

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P
     R
          Q
          P
     Q -› P
R -› (Q -› P)

Premise
* A
* A
* 1 Reit
* 3-4 -›I
2-5 -›I


You should now:

Go to Lesson 2

        Let us now look at another example. We will show that P -› ~P, ~(Q v ~R) I- S -› (~P & R).

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P -› ~P
~(Q v ~R)
     S
          P
          ~P
          P & ~P
     ~P
          ~R
          Q v ~R
          (Q v ~R) & ~(Q v ~R)
     ~~R
     R
     ~P & R
S -› (~P & R)

Premise
Premise
* A
* A
* 1, 4 -›E
* 4, 5 &I
* 3-6 ~I
* A
* 8 vI
* 2, 9 &I
* 8-10 ~I
* 11 ~~E
* 7,12 &I
3-13 -›I


Notice that in completing this derivation we have combined tactics used in preceding examples. We wished to derive a conditional, so we set up a conditional introduction. The consequent of this conditional was a conjunction, so we obtained the conjuncts and then used &I. You should now:

Go to Lesson 3

Before continuing to the next tutorial or the exercises let us try one more derivation. You should now:

Go to Lesson 4

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