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Reasoning “backwards” in natural deduction: a one-proof tutorial. PHIL 120: Introduction to Logic Tyler Hildebrand. Introduction.
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Reasoning “backwards” in natural deduction: a one-proof tutorial PHIL 120: Introduction to Logic Tyler Hildebrand
Introduction • Many students are able to reason through simple natural deduction proofs using a “top down” or “forwards” strategy, in which you start the proof by applying inference rules to the premises (and other randomly chosen assumptions) until you arrive at the conclusion. • Unfortunately, there are many sequents for which the “forwards” strategy is unlikely to yield a correct proof. This tutorial will walk you through a different strategy: the “bottom up” or “backwards” strategy, in which you start by examining the conclusion and determining which formula must occur on the line above it, and then repeating that reasoning for subsequent lines as you work to fill in the proof from the conclusion up.
A Definition • The target formula is whatever line you’re attempting to derive. • Thus the initial target formula of a proof is always the conclusion (occurring on line c). • Once you have selected a strategy for deriving the conclusion from one or more formulae on lines above c, those formulae become new target formulae, and so on.
Example 1 (1) A(BC) Pr 1 ( ) (AB)(AC) We begin with our premise, written on line (1), and our conclusion, written at the bottom of the proof without any label. Since we are just starting, the conclusion is our target formula. Examining our conclusion and our premise, we see that the conclusion does not occur as the subformula in the premise.
Example 1 (1) A(BC) Pr 1 ( ) (AB)(AC) Thus we move to the next step. Note that the conclusion is a conditional (that is, that its main connective is ). Accordingly, we will attempt to derive the conclusion using the introduction rule for that connective: I.
Example 1 (1) A(BC) Pr 1 ( ) (AB)(AC) Thus we write ‘I’ to the right of the conclusion, indicating the strategy we wish to employ. Now that we’ve selected a strategy for deriving our target formula (namely, using I), we can fill in some other information into our proof. This information is provided by the rule itself. I
Schematic form of I j (j) p Assp . . a1,…,an (k) q . . {a1,…,an}/j (m) pqj,k I
Example First, we must assume the antecedent, AB, of the conditional at the top of our proof. IT IS VERY IMPORTANT THAT WE WRITE THE ANTECEDENT IN ITS ENTIRETY. ‘A’ would not suffice. 1 (1) A(BC) Pr 1 ( ) (AB)(AC) I 2 (2) AB Assp ( ) AC Second, we must write the consequent of the conditional, (AC), on the line above our target formula. This will become our new target formula.
Example One last thing to do before we move on to consider our new subformula: we should fill in line numbers for the justification of the step we’ve just completed. (It’s easier to do this now than later.) We now know that the I step on the last line depends on line 2, so let’s write that down. 1 (1) A(BC) Pr 2 (2) AB Assp ( ) AC 1 ( ) (AB)(AC) I 2,__ We can now restart this process of reasoning for the next target formula: AC.
Example 1 (1) A(BC) Pr 2 (2) AB Assp ( ) AC 1 ( ) (AB)(AC) 2,__I Does AC occur as a subformula in any of the premises, assumptions, or formulae derived from the premises or assumptions? No. However, the main connective of our target formula is , so we should (once again) consider deriving it using I. (You’ll come to love this rule, because it provides you with another assumption to work with.)
Example 1 (1) A(BC) Pr 2 (2) AB Assp ( ) AC 1 ( ) (AB)(AC) 2,__I Begin by writing the antecedent, A, of our target formula on the next available line at the top of the proof as an assumption, depending on itself. 3 (3) A Assp ( ) C Next, write the consequent, C, of our target formula on the line above our target formula. 3,__I Finally, fill in the line numbers where possible.
Example 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp ( ) C ( ) AC 3,__I 1 ( ) (AB)(AC) 2,__I At this point you may be able to see how to complete the proof by reasoning “forwards.” Resist that urge. It is important to understand how to complete a proof working backwards from start to finish, because you may encounter proofs for which the forwards strategy never becomes intuitive.
Example 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp ( ) C ( ) AC 3,__I 1 ( ) (AB)(AC) 2,__I Anyway, our new target formula is C, so we repeat our reasoning process. An inspection of the premises indicates that C occurs as a subformula of line (1). However, there is no inference rule that will allow us to go directly from the formula at line (1) to C, so we’ll have to break the formula on line (1) into smaller components.
Example 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp ( ) C ( ) AC 3,__I 1 ( ) (AB)(AC) 2,__I The key move is to note that if we were to obtain BC on its own line, the we could try to obtain B on its own line, and thereby obtain C from an application of E to the aforementioned formulae. Thus let us suppose that we will obtain C by an application of E. E
Example 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp ( ) C E ( ) AC 3,__I 1 ( ) (AB)(AC) 2,__I Now we need to add lines above our target formula as required by our strategy (namely, the decision to obtain C by E). To be perfectly clear, then, let’s have a quick look at the schematic form of E.
Schematic form of E a1,…,an (j) pq . . b1,…,bu (k) p . . a1,…,an, b1,…,bu (m) q j,k E
Example 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp ( ) C E ( ) AC 3,__I 1 ( ) (AB)(AC) 2,__I We need to add two lines: One is BC, as we already know. ( ) BC The other is B, since that is the formula required to derive C from BC by E. ( ) B Note that their order doesn’t matter. They just need to appear on available lines at the bottom of the proof.
Example 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp ( ) BC ( ) B ( ) C E ( ) AC 3,__I 1 ( ) (AB)(AC) 2,__I Now we have two target formula. We’ll consider them one at a time. (Don’t forget to consider both!)
Example First, let’s consider B. An inspection of the premises assumptions reveals that B is a subformula of line (2). Thus we should consider deriving it using the E. 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp ( ) BC ( ) B ( ) C E ( ) AC 3,__I 1 ( ) (AB)(AC) 2,__I E An application of E requires us to have A, the antecedent of AB, occurring on its own line. We already have that on line (3)!Thus we can fill in lots of information.
Example First, let’s write the line numbers in the justification column that correspond to our conditional and its antecedent. 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp ( ) BC (?) B ( ) C E ( ) AC 3,__I 1 ( ) (AB)(AC) 2,__I 2,3 2,3 E Second, let’s fill in the leftmost column as well. Since line (2) depends only on itself, and line (3) depends only on itself, our formula B on line (?) depends on lines (2) and (3).
Example Note that, because we didn’t have to introduce a new target formula—we already had everything we needed to derive the target formula B on line (?)—we are done with the current line of reasoning. 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp ( ) BC 2,3 (?) B 2,3 E ( ) C E ( ) AC 3,__I 1 ( ) (AB)(AC) 2,__I However, we are NOT done with the proof, since we have another target formula, BC. We need to keep applying our strategy to ALL target formulae until each line of reasoning is fully resolved.
Example With that in mind, how should we derive BC? An inspection of the premises, assumptions, and derived lines reveals that BC is a subformula of the formula on line (1). Thus we should attempt to derive it using E on line (1). 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp ( ) BC 2,3 () B 2,3 E ( ) C E ( ) AC 3,__I 1 ( ) (AB)(AC) 2,__I Fortunately, we already have the antecedent of the conditional on its own line: A on line (3). Thus we can fill in the proof just as we did in the preceding step.
Example Our target formula on line (?) is derived from lines (1) and (3) by E. Since lines (1) and (3) depend only on themselves, we write 1,3 in the leftmost column of that line to indicate that our target formula depends on the formulae on those two lines. 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp (?) BC 2,3 ( ) B 2,3 E ( ) C E ( ) AC 3,__I 1 ( ) (AB)(AC) 2,__I 1,3 1,3 E We have no remaining target formulae! To finish the proof, we simply need to fill in the relevant line numbers, etc.
Example First, let’s fill in the line numbers. 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp 1,3 (4) BC 1,3 E 2,3 (5) B 2,3 E (6) C E (7) AC 3,__I 1 (8) (AB)(AC) 2,__I Second, let’s work from top to bottom, filling in the justification column and the leftmost column line by line. Lines (1) through (5) are already complete, so we just need to fill in (6) through (8).
Example Line (6) was derived by E, so it must have been derived from lines (4) and (5). We indicate this on the right. 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp 1,3 (4) BC 1,3 E 2,3 (5) B 2,3 E (6) C (7) AC 3,__I 1 (8) (AB)(AC) 2,__I (6) depends on everything that lines (4) and (5) depend on. (4) depends on (1) and (3); (5) depends on (2) and (3); therefore, (6) depends on (1), (2), and (3). We write this on the left. 4,5 E 1,2,3
Example We used I to derive line (7). If you’ve forgotten which assumption was used, you can simply see which assumption is the antecedent of the formula on line (7). 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp 1,3 (4) BC 1,3 E 2,3 (5) B 2,3 E 1,2,3 (6) C 4,5 E 1,2 (7) AC 3,6 I 1 (8) (AB)(AC) 2,__I The justification for (7) is thus lines (3) and (6). (7) depends on everything (6) depends on except the original assumption (3).
Example The same type of reasoning is then employed to fill in line (8). 1 (1) A(BC) Pr 2 (2) AB Assp 3 (3) A Assp 1,3 (4) BC 1,3 E 2,3 (5) B 2,3 E 1,2,3 (6) C 4,5 E 1,2 (7) AC 3,6 I 1 (8) (AB)(AC) 2,7 I We’re done!
Description of Strategy: Step 1 • Step 1: Examine the target formula and the premises; determine whether the target formula occurs as a subformula in any of the premises or lines already derived from the premises and determine the main connective of the target formula. • Example: The target formula is P and one of the premises is (Q v R) P.
Description of Strategy: Step 2 • Step 2: If the target formula is a subformula of one of the premises then consider obtaining it using the elimination rule for the relevant connective of the formula in the premise. • Example: The target formula is P and one of the premises is (Q v R) P. We’d consider obtaining P by E. This would require us to derive Q v R on some line higher up in the proof.
Description of Strategy: Step 3 • If Step 2 doesn’t apply, consider obtaining the target formula using the introduction rule for the main connective of the target formula. • Example: If the target formula is AB, consider deriving it with I by assuming A and attempting to derive B on the line above the target formula.
Description of Strategy: Step 4 • If Step 3 doesn’t apply, consider obtaining the target formula by reductio. • Example 1: The target is a negation, say, ~P. Assume P at the top of your proof and write the absurdity on the line above ~P. • Example 2: The target is not a negation, say, A&B. Derive A&B by DN, writing ~~(A&B) on the line above. Then attempt to derive that formula as in Example 1.
