The Indirect Method Lesson 1: Proofs of Common Forms
In the last course we learned how to use truth-tables to test argument forms (otherwise known as “sequents”) for validity. Truth-tables are a very useful tool for the evaluation of argument forms. However, as we saw in the last course, truth-tables can become unwieldy when applied to argument forms containing more than two simple sentences.
If we only have two simple sentences to deal with, our truth-table will only be four lines deep. However, with each additional simple sentence the number of lines doubles. This means that if we have four simple sentences our truth-table will be thirty-two lines deep! And with thirty two lines to worry about, there is a significant chance that a mere mortal such as you or I will make a mistake. This is where the indirect method comes in handy. Using the indirect method we can test the validity of sequents containing three, four, five or more simple sentences quickly and efficiently.
The indirect method works according to the principle that if an assumption leads to contradiction, then that assumption must be false.Thus, when applying the indirect method to test for validity we start by assuming that our sequent is invalid. If this assumption leads us into contradiction, we know that our assumption is false, and that our sequent is in fact valid. However, if our assumption does not lead to contradiction then we know that our sequent is invalid.
In this lesson you will be introduced to the indirect method and will be required to apply this method first to the simplest versions of the familiar forms MP, MT, AC, DA and HS and second to some more complex versions of these forms.
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.
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 WFF 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 unassigned 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 formulae 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: Modus Ponens
Set the argument out horizontally:
| (P | → | Q) | 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.
| (P | → | Q) | P | ⊧ | Q |
| T | T | F | T | F |
The proof is now complete (all simple sentences and logical operators have been assigned truth-values) so we can skip rules 2-3 and go straight to rule 4.
| (P | → | Q) | P | ⊧ | Q |
| T | T | F | T | F |
As we can see, there is a contradiction in the first premise (If the antecedent of a conditional is true, and its consequent is false, then the truth-value of the entire conditional should be F, and not T). 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: Affirming the Consequent
Let’s consider an invalid argument, affirming the consequent:
| (P | → | Q) | 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) | Q | ⊧ | P |
| F | T | T | T | F |
The proof is now complete (all simple sentences and logical operators have been assigned truth-values) so we can now go ahead and apply Rule 5.
| (P | → | Q) | Q | ⊧ | P |
| F | T | T | T | F |
As we can see, there is no contradiction. Given that our original assumption (that the premises are true and the conclusion is false) has not led us into contradiction, it must 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 invalid.
EXAMPLE 3: Hypothetical Syllogism
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 | → | R) | ⊧ | (P | → | R) | ||
| T | T | F |
The application of Rule 1 has left three formulae unassigned, 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.
The three unassigned WFF’s are all simple sentences, and thus, are all of equal breadth. Thus, we will start with the leftmost unassigned WFF, “P”. As “P” is a simple sentences, “P” does not contain any narrower formulae, and this cannot have its TV internally determined. Thus, we must look for a Externally Determining WFF for “P”.
P appears in the context of two assigned WFF’s. Both the first premise, “(P → Q)”, and the conclusion, “(P → R)” contain “P”. ” However, the truth of “(P → Q)” does not entail either the truth or falsity of “P”. Thus, “(P → Q)” is not an EDF for “P”. Happily, the falsity of “(P → R)” is sufficient to entail the truth of “P” (if “(P → R)” is false then “P” must be true). Thus, we will go ahead as assign the TV “T” to all tokens of “P” as follows:
| (P | → | Q) | (Q | → | R) | ⊧ | (P | → | R) | ||
| T | T | T | T | F |
Now we can apply Rule 2 to “Q”, which is now the leftmost and equally broadest of the two remaining unassigned WFF’s. “Q” is contained in two equally broad WFF’s, “(P → Q)” and “(Q → R)”. As “(P → Q)” is to the left of “(P → Q)”, we start by checking to see if the TV of “(P → Q)” entails the TV of “Q”. As it happens, we are in luck. Given that “(P → Q)” is true, and “P” is true, “Q” must also be true. Thus, we can go ahead and assign a TV of “T” to all tokens of “Q” in our proof as follows:
| (P | → | Q) | (Q | → | R) | ⊧ | (P | → | R) | ||
| T | T | T | T | T | T | F |
So, now the only remaining unassigned WFF is “R”. “R” is contained in two WFF’s, “(Q → R)” and “(P → R)”. As the former is to the left of the latter, we will start by checking if the former WFF is an EDF for “R”. As it happens, the truth of “(Q → R)” together with the truth of “Q” entails that “R” must also be true. Thus, we will go ahead and assign the TV of “T” to all tokens of “R” as follows:
| (P | → | Q) | (Q | → | R) | ⊧ | (P | → | R) | ||
| T | T | T | T | T | T | T | F | T |
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 | → | R) | ⊧ | (P | → | R) | ||
| T | T | T | T | T | T | T | F | T |
As we can see, there is a contradiction in the conclusion (If both the antecedent and the consequent of a conditional are true then the truth-value of the entire conditional should be T, and not F). 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 4: A Complex Version of Modus Tollens
Let’s now look at a more complex argument, this time a complex version of modus tollens:
| (P | → | (Q | v | R)) | ~ | (Q | v | R) | ⊧ | ~ | 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 | v | R)) | ~ | (Q | v | R) | ⊧ | ~ | P | ||
| 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 v R)”, which appears in both the first and second premises. Neither of the simple sentences contained with “(Q v R)” is assigned at this point, so “(Q v R)”s TV is not internally determined. However, the TV of “(Q v R)” is externally determined by its inclusion within the assigned WFF “~(Q v R)”. As “~(Q v R)” is true, “(Q v R)” must be false. Thus, we will go ahead and assign.
| (P | → | (Q | v | R)) | ~ | (Q | v | R) | ⊧ | ~ | P | ||
| T | F | T | F | F |
Now we can repeat the application of Rule 2 to all three remaining unassigned WFF’s “P”, “Q” and “R”.
The narrowest assigned WFF in which “P” appears is “~P” (i.e. the conclusion). What is more, the assigned TV of “~P”, “F”, is sufficient to determine the TV of P: if “~P” is false then “P” must be . Thus, we can go ahead and assign a TV of “T” to both tokens of “P” as follows:
| (P | → | (Q | v | R)) | ~ | (Q | v | R) | ⊧ | ~ | P | ||
| T | T | F | T | F | F | T |
Next, we turn to “Q” and “R”. For an inclusive disjunction to be false, both of its disjuncts must be false. Thus, the falsity of “(Q v R)” entails the falsity of both “Q” and “R”. Thus, we can go ahead and assign the TV “F” to all tokens and “Q” and “R” as follows:
| (P | → | (Q | v | R)) | ~ | (Q | v | R) | ⊧ | ~ | P | ||
| T | T | F | F | F | T | F | F | F | 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 | v | R)) | ~ | (Q | v | R) | ⊧ | ~ | P | ||
| T | T | F | F | F | T | F | F | F | 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.
EXAMPLE 4: A Complex Version of Denying the Antecedent
Now let’s see how the procedure works with another invalid argument: this time a complex version of denying the antecedent.
| (P | → | (Q | & | R)) | ~ | P | ⊧ | ~ | (Q | & | R) | ||
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)) | ~ | P | ⊧ | ~ | (Q | & | R) | ||
| 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 remaining unassigned WFF in our proof is “(Q & R)” and the narrowest WFF in which this WFF appears is the conclusion, “~(Q & R)”. What is more, the TV of “~(Q & R)” is sufficient to determine the TV of “(Q & R)”: if “~(Q & R)” is false then “(Q & R)” must be true. Thus, we can go ahead and assign a TV of T to both tokens of “(Q & R)”.
| (P | → | (Q | & | R)) | ~ | P | ⊧ | ~ | (Q | & | R) | ||
| T | T | T | F | T |
Now we can repeat the application of Rule 2 to all three remaining unassigned WFF’s “P”, “Q” and “R”. “The narrowest assigned WFF in which “P” appears is the second premise, “~P”. What is more, the truth of “~P” is sufficient to determine the TV of P: if “~P” is true then “P” must be false. Thus, we can go ahead and assign a TV of F to both tokens of “P” as follows:
| (P | → | (Q | & | R)) | ~ | P | ⊧ | ~ | (Q | & | R) | ||
| F | T | T | T | F | F | T |
Rule 2 also allows us to assign TV’s to “Q” and “R”. For a conjunction to be true, both of its conjuncts must be true. Thus, the truth of “(Q & R)” entails the truth of both “Q” and “R”. Thus, we can go ahead and assign the TV T to all tokens and “Q” and “R” as follows:
| (P | → | (Q | & | R)) | ~ | P | ⊧ | ~ | (Q | & | R) | ||
| F | T | T | T | T | T | F | F | T | T | T |
The proof is now complete (all simple sentences and logical operators have been assigned truth-values) so we can now apply Rule 4. As we can see, there is no contradiction within 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 possible for the conclusion of our argument to be false at the same time that its premises are true. Thus, our argument must be invalid.
The videos contain more explanation and examples: