The Indirect Method Lesson 2: Proofs of Uncommon Forms
In the last lesson we learned how to use the Indirect Method to test some common argument forms (and complex variations on those forms) for validity. In this lesson we will deploy to Indirect Method to test some uncommon forms for validity.
Here again are our key definitions:
Definitions:
TV = Truth-Value
WFF = Well-Formed Formula
Assigned WFF = A well-formed formula whose main operator has already been assigned a truth-value.
Unassigned WFF = A well-formed formula whose main operator has not yet been assigned a truth-value.
X’s TV is internally determined = The TV of one or more assigned WFF’s that appear within X are sufficient to entail X’s TV.
X’s TV is externally determined = The TV of one or more of the assigned formulae in which X appears (together with the TV’s of any other assigned WFF’s that appear within those formulae) is sufficient to entail the TV of X.
Externally Determining Formula (EDF) = X is an EDF for Y if and only if Y appears within X and X’s assigned TV entails the TV of Y.
And here again are the rules that will follow when applying the Indirect Method:
The Indirect Method Procedure
(1) Assign T to the main operator of each premise and F to the main operator of the conclusion, and the same TV to any token WFF’s of the same AFF type.
(2) Is there a unique broadest unassigned WFF within the proof? If yes, we will call this WFF “X”. If not, we will call the leftmost broadest WFF “X”.
Q1: Is X’s TV internally determined?
Q2: Is X’s TV externally determined?
If the answer to Q1 is yes, assign the internally determined TV to all tokens of X.
If the answer to Q1 is no, but the answer to Q2 is yes, assign to X the TV entailed by X’s narrowest externally determining EDF.
If the answer to both questions is “No”, move onto the next broadest unassigned formula.
(3) If one or more WFF’s remain unassigned even after the repeated application of Rule 2, assign the TV “T” to all tokens of the leftmost unassigned simple sentence then continue to apply Rule 2.
(4) Once all formula have been assigned TV’s, check to see if there is a contradiction within the proof. If there is a contradiction within the proof, the sequent is valid. If there is no contradiction within the proof, the sequent is invalid.
EXAMPLE 1
Set the argument out horizontally:
| (~ | Q | x∨ | P) | P | ⊧ | Q | ||
Now, apply (Rule 1): Assign T to the main operator of each premise and F to the main operator of the conclusion, and the same TV to any token WFF’s of the same AFF type.
| (~ | Q | x∨ | P) | P | ⊧ | Q | ||
| F | T | T | T | F |
As there is one WFF that remains unassigned (i.e. “~Q”), we will need to apply Rule 2: Is there a unique broadest unassigned WFF within the proof? If yes, we will call this WFF “X”. If not, we will call the leftmost broadest WFF “X”.
Q1: Is X’s TV internally determined?
Q2: Is X’s TV externally determined?
If the answer to Q1 is yes, assign the internally determined TV to all tokens of X.
If the answer to Q1 is no, but the answer to Q2 is yes, assign to X the TV entailed by X’s narrowest externally determining EDF.
If the answer to both questions is “No”, move onto the next broadest unassigned formula.
As “Q” has already been assigned “F”, the TV of “~Q” is internally determined: if “Q” is “F”, then :~Q” must be “T”. So, we will go ahead and complete the proof as follows:
| (~ | Q | x∨ | P) | P | ⊧ | Q | ||
| T | F | T | T | T | F |
As we can see, there is a contradiction in the first premise (If both disjuncts of an exclusive disjunction are true, then the entire disjunction is false. Given that our original assumption (that the premises are true and the conclusion is false) has led us into contradiction, that original assumption must be false: it must be impossible for the conclusion of our argument to be false at the same time that its premises are true. Thus, our argument must be valid.
EXAMPLE 2
Let’s consider a slightly more complex argument:
| (Q | ↔ | ~ | P) | ~ | P | ⊧ | Q | ||
Now, apply (Rule 1): Assign T to the main operator of each premise and F to the main operator of the conclusion, and the same TV to any token WFF’s of the same AFF type.
| (Q | ↔ | ~ | P) | ~ | P | ⊧ | Q | ||
| F | T | T | T | F |
As there is still one unassigned WFF (i.e. “P”), we will need to apply Rule 2: Is there a unique broadest unassigned WFF within the proof? If yes, we will call this WFF “X”. If not, we will call the leftmost broadest WFF “X”.
Q1: Is X’s TV internally determined?
Q2: Is X’s TV externally determined?
If the answer to Q1 is yes, assign the internally determined TV to all tokens of X.
If the answer to Q1 is no, but the answer to Q2 is yes, assign to X the TV entailed by X’s narrowest externally determining EDF.
If the answer to both questions is “No”, move onto the next broadest unassigned formula.
As “P” is only contained in one WFF, “~P”, and “~P” is “T”, “P” must be “F”. So, we will go ahead and complete the proof as follows:
| (Q | ↔ | ~ | P) | ~ | P | ⊧ | Q | ||
| F | T | T | F | T | F | F |
As we can see, there is again a contradiction within our proof. Given that our original assumption (that the premises are true and the conclusion is false) has led us into contradiction, it must not be possible for the conclusion of our argument to be false at the same time that its premises are true. Thus, our argument must be valid.
EXAMPLE 3
| ~ | (P | & | ~ | Q) | ~ | Q | ⊧ | P | ||
Set the argument out horizontally then apply (Rule 1): Assign T to the main operator of each premise and F to the main operator of the conclusion, and the same TV to any token WFF’s of the same AFF type.
| ~ | (P | & | ~ | Q) | ~ | Q | ⊧ | P | ||
| T | F | T | T | F |
The application of Rule 1 has left two formulae unassigned (“Q” and “(P & ~Q)”) so we must apply Rule 2:
(Rule 2) Is there a unique broadest unassigned WFF within the proof? If yes, we will call this WFF “X”. If not, we will call the leftmost broadest WFF “X”.
Q1: Is X’s TV internally determined?
Q2: Is X’s TV externally determined?
If the answer to Q1 is yes, assign the internally determined TV to all tokens of X.
If the answer to Q1 is no, but the answer to Q2 is yes, assign to X the TV entailed by X’s narrowest externally determining EDF.
If the answer to both questions is “No”, move onto the next broadest unassigned formula.
As “(P & ~Q)” is broader than “Q” we will start with the former. The TV of “(P & ~Q)” is internally determined: as “P” has been assigned “F”, the TV of “(P & ~Q)” must also be “F”. Thus, we will go ahead as assign the TV “F” to all tokens of “(P & ~Q)” as follows:
| ~ | (P | & | ~ | Q) | ~ | Q | ⊧ | P | ||
| T | F | F | T | T | F |
Now we can apply Rule 2 to “Q”, which is now the only remaining unassigned WFF. “Q” is contained in “~Q” which has already been assigned the TV “T”. Further, if ~Q” is “T” then “Q” must be “F”. Thus, we can go ahead and assign a TV of “F” to all tokens of “Q” as follows:
| ~ | (P | & | ~ | Q) | ~ | Q | ⊧ | P | ||
| T | F | F | T | F | T | F | F |
The proof is now complete (all simple sentences and logical operators have been assigned truth-values) so we can skip rules 3-4 and go straight to rule 4.
| ~ | (P | & | ~ | Q) | ~ | Q | ⊧ | P | ||
| T | F | F | T | F | T | F | F |
As we can see, there is no contradiction in the proof. Given that our original assumption (that the premises are true and the conclusion is false) has not led us into contradiction, it must be impossible for the conclusion of our argument to be false at the same time that its premises are true. Thus, our argument must be invalid.
EXAMPLE 4
Let’s now look at an even more complex argument:
| (P | → | (~ | Q | & | R)) | Q | ⊧ | ~ | P | ||
Now, apply (Rule 1): Assign T to the main operator of each premise and F to the main operator of the conclusion, and the same TV to any token WFF’s of the same AFF type.
| (P | → | (~ | Q | & | R)) | Q | ⊧ | ~ | P | ||
| T | T | T | F |
Now, let’s apply Rule 2: Is there a unique broadest unassigned WFF within the proof? If yes, we will call this WFF “X”. If not, we will call the leftmost broadest WFF “X”.
Q1: Is X’s TV internally determined?
Q2: Is X’s TV externally determined?
If the answer to Q1 is yes, assign the internally determined TV to all tokens of X.
If the answer to Q1 is no, but the answer to Q2 is yes, assign to X the TV entailed by X’s narrowest externally determining EDF.
If the answer to both questions is “No”, move onto the next broadest unassigned formula.
The broadest unassigned WFF is “(~Q & R)”, which appears in the first premise. Neither “~Q” or the nor “R” have been assigned at this point, so “(~Q & R)”s TV is not internally determined. Further, the TV of “(Q v R)” is externally determined by its inclusion within the true WFF “(P → (~Q & R))”. Thus, we will turn our attention to the next broadest WFF, “~Q”.
As “Q” has already been assigned “T”, “~Q”‘s TV is internally determined: if “~Q” is “T” then “~Q” must be “F”. So, we will go ahead and assign “~Q” the TV “F” as follows:
| (P | → | (~ | Q | & | R)) | Q | ⊧ | ~ | P | ||
| T | F | T | T | F |
Now that “~Q” has been assigned the TV “F”, the TV if “(~Q & R)” is internally determined: if “~Q” is “F” then “(~Q & R)” must also be assigned “F” as follows:
| (P | → | (~ | Q | & | R)) | Q | ⊧ | ~ | P | ||
| T | F | T | F | T | F |
So, now the only two remaining unassigned WFF’s are “P” and “R”. The narrowest determining WFF in which “P” appears is “~P” in the conclusion: given that “~P” is false, “P” must be assigned “T” as follows:
| (P | → | (~ | Q | & | R)) | Q | ⊧ | ~ | P | ||
| T | T | F | T | F | T | F | T |
Finally, we turn to “R”. For conjunction to be false, only one conjunct need be false. Thus, the falsity of “(~Q & R)” entails neither the truth nor the falsity of R. This is where Rule 3 comes into play: as the TV of R is not determined by the TV’s of any of the other WFF’s in the proof, we go ahead and assign “R” the TV “T” as follows:
| (P | → | (~ | Q | & | R)) | Q | ⊧ | ~ | P | ||
| T | T | F | T | F | T | T | F | T |
The proof is now complete (all simple sentences and logical operators have been assigned truth-values) so we can now apply Rule 4.
| (P | → | (~ | Q | & | R)) | Q | ⊧ | ~ | P | ||
| T | T | F | T | F | T | T | F | T |
As we can see, there is a contradiction in Premise 1. If the antecedent of a conditional is true and its conclusion is false (as is the case in Premise 1 of our proof) then the entire conditional must be false. However, the truth-value assigned to the main operator of Premise 1 in our proof is “T”. Given that our original assumption (that the premises are true and the conclusion is false) has led us into contradiction, it must not be possible for the conclusion of our argument to be false at the same time that its premises are true. Thus, our argument must be valid.
The videos contain more explanation and examples: