

Sample sentences AND operator IF/THEN operator NOT operator OR operator XOR operator Chain Rule Conjunctive Addition Contrapositive DeMorgan's Law Disjunctive Addition Disjunctive Inference Disjunctive Infer. (XOR) Double Negation Modus Ponens Modus Tollens Mutual Exclusion Simplification 2step 3step 4step 5step or more Bad Argument 
Validity Proof Problems (2 steps) The problems below can all be solved in two steps.
Problem 1 1. G ^ K Using the premises in 1 and 2, prove that the conclusion G ^ H is true. We see that G is part of premise 1, and H is part of premise 2, and we want to bring them together in order to prove that G ^ H is true. We can separate G out by using Simplification. 3. G 1 Simplification We have written our new step 3 as a result of applying Simplification to premise 1. Now that we have G separated, and knowing that H is given, we can join them by using Conjunctive Addition. 4. G ^ H 3,2 Conjunctive Addition This line reads as follows: 4 is the new step number; G ^ H is the result of applying Conjunctive Addition to premises 3 and 2. We have gotten to the conclusion. 1. G ^ K
Problem 2 1. R > S If we look at premises 2 and 3, we can see that it is possible to apply Modus Ponens to them, in order to isolate R. We know that we are going to need R in order to obtain S from premise 1, which is our conclusion. 4. R 3,2 Modus Ponens Now that we have R, we can apply Modus Ponens to premise 1. Here is what happens: 5. S 1,4 Modus Ponens We have gotten to our conclusion, S. Let's review our steps. 1. R > S There is another way to solve this problem. Let's restate it. 1. R > S If we look carefully at premises 3 and 1, we will realize that we can apply the Chain Rule here. 4. U > S 3,1 Chain Rule Now we can apply Modus Ponens to the result. 5. S 4,2 Modus Ponens We've reached the same conclusion, with the same number of steps. 1. R > S
Problem 3 1. ~G > L Let's look at premises 2 and 3. We can apply Modus Tollens here. 4. ~G 2,3 Modus Tollens Now that we have ~G, we can look at premise 1 and realize that by applying Modus Ponens we can get L. 5. L 1,4 Modus Ponens We are done. 1. ~G > L
Problem 4 1. D v E We see right away that we can apply Disjunctive Inference to premises 1 and 2. D is true because E is false. 3. D 1,2 Disjunctive Inference Now we can add anything to D using Disjunctive Addition. For example, we can add F: 4. D v F 3 Disjunctive Addition We've reached the conclusion. 1. D v E
Problem 5 1. K > L Look closely at premises 1 and 3. This looks like a perfect example of Modus Tollens. 4. ~K 1,3 Modus Tollens We have a disjunction in premise 2, and we already know that K is not true. By Disjunctive Inference, we know that J must be true. 5. J 2,4 Disjunctive Inference We've reached the conclusion. 1. K > L
Problem 6 1. G XOR H Premise 2 states that H is true. Looking at premise 1, we realize that G has to be false  that is the property of the XOR operator, and an example of Mutual Exclusion. 4. ~G 1,2 Mutual Exclusion Now that we have ~G, we can use Disjunctive Inference with XOR on premise 3 to prove that J is true. 5. J 3,4 Disjunctive Inference with XOR We have gotten to the conclusion. 1. G XOR H
