Subsection 3.5.7 Conditional Disjunction Proof Problem
Prove: q ∨ p
¬ p
¬( q ∧ s )
∴ s → r
(Hint: There are two reasonable and quite different ways to do this proof. One involves Conditionalization. We’ll say more about that soon. So you probably want to do this one without Conditionalization. Instead, you may want to use both Conditional Disjunction and Addition. Recall that Addition lets you introduce a variable “out of thin air”. Look at the statement of this problem to see why you might need to do that.)
You should do this proof yourself.
You can also watch our video, which will outline our strategy for doing this.
Exercises Exercises
Exercise Group.
1.
1. Prove: R → W
W → S
S → C
R
C → I
∴ I
(Hint: This one is easy. Focus on how you can use Modus Ponens.)
[1] \(R \rightarrow W \) Premise
[2] \(W \rightarrow S \) Premise
[3] \(S \rightarrow C \) Premise
[4] \(R \) Premise
[5] \(C \rightarrow I \) Premise
[6] W Modus Ponens [1], [4]
[7] S Modus Ponens [2], [6]
[8] C Modus Ponens [3], [7]
[9] I Modus Ponens [5], [8]
2.
Prove: R → W
W → S
S → C
R
W → Y
R ∧ Y
(Hint: What rule lets you create a conjunction out of two or more statements you already have?)
questionId: PsQs10
problemType: gradeLogicProof
questionTitle: A Simple Direct Proof
questionDisplayText: We need to prove two things, R and Y.
problemGoal: R Y
initialProblemState: the five premises
hints: You need to conclude with a rule that lets you combine multiple statements into a conjoined one.
[1] \(R \rightarrow W \) Premise
[2] W S Premise
[3] S C Premise
[4] R Premise
[5] W Y Premise
[6] W Modus Ponens 1, 4
[7] Y Modus Ponens 5, 6
[8] W Y Conjunction 4, 7