- #1

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Prove that this is a valid argument using reductio ad absurdum

[tex]

1. (A \supset (B \bullet C)) [/tex]

[tex] 2. (B \supset (A \bullet C)) [/tex]

[tex] Therefore, ((A \vee B) \supset C)

[/tex]

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In summary: Conditional Elimination 6 || ~~C 5 Conjunction Elimination 7 || ~~B AssumptionStraightforward2 | (B > (A & C)) Premise|---------------------------- 3 || ~~B v ~C

- #1

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Prove that this is a valid argument using reductio ad absurdum

[tex]

1. (A \supset (B \bullet C)) [/tex]

[tex] 2. (B \supset (A \bullet C)) [/tex]

[tex] Therefore, ((A \vee B) \supset C)

[/tex]

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- #2

Gold Member

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ASkomatth said:Since this forum seems a little slow

I don't know what rules I can use (?), so if you want something, just ask. As much as I love it, [itex]\LaTeX[/itex] is taking forever, so: ~ = NOT; & = AND; v = OR; -> = IMPLIES.Prove that this is a valid argument using reductio ad absurdum

[tex]

1. (A \supset (B \bullet C)) [/tex]

[tex] 2. (B \supset (A \bullet C)) [/tex]

[tex] Therefore, ((A \vee B) \supset C)

[/tex]

1) A -> (B & C) [premise]

2) B -> (A & C) [premise]

3) ~A v (B & C) [1]

4) (~A v B) & (~A v C) [3]

5) ~A v C [4]

6) ~B v (A & C) [2]

7) (~B v A) & (~B v C) [6]

8) ~B v A [7]

9)) (A v B) & ~C [assumption]

10)) A v B [9]

11)) ~C [9]

12)) ~A [5, 11]

13)) ~B [8, 12]

14)) B [10, 12]

15)) B & ~B [13, 14]

16) ~((A v B) & ~C) [9, 15, reductio]

17) ~(A v B) v ~~C [16]

18) ~(A v B) v C [17]

19) (A v B) -> C [18, QED]

Wow, I must be getting rusty - that seems too long. The thing to notice is that A <-> B. Maybe I should have used that. Meh. I don't think I made any mistakes, at least. This isn't homework, is it?

Last edited:

- #3

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Using Reductio ad Absurdum? Okay, but the straightforward proof is much faster. I'll give both proofs:

__Straightforward__
__A different way__
__Reductio__

Code:

```
1 | (A > (B & C)) Premise
2 | (B > (A & C)) Premise
|----------------------------
3 || (A v B) Assumption
||---------------------------
4 ||| A Assumption
|||--------------------------
5 ||| (B & C) 1, 4 Conditional Elimination
6 ||| C 5 Conjunction Elimination
||
7 ||| B Assumption
|||--------------------------
8 ||| (A & C) 2, 7 Conditional Elimination
9 ||| C 8 Conjunction Elimination
10 || C 3, 4-6, 7-9 Disjunction Elimination
11 | ((A v B) > C) 3-10 Conditional Introduction
```

Code:

```
1 | (A > (B & C)) Premise
2 | (B > (A & C)) Premise
|----------------------------
3 || ~C Assumption
||---------------------------
4 || ~C v ~B 3 Disjunction Introduction
5 || ~(B & C) 4 DeMorgan's, Commutativity
6 || ~A 1, 5 Modus Tollens
7 || ~C v ~A 3 Disjunction Introduction
8 || ~(A & C) 7 DeMorgan's, Commutativity
9 || ~B 2, 8 Modus Tollens
10 || (~A & ~B) 6, 9 Conjunction Introduction
11 || ~(A v B) 10 DeMorgan's
12 | (~C > ~(A v B)) 3-11 Conditional Introduction
13 | ((A v B) > C) 12 Transposition
```

Code:

```
1 | (A > (B & C)) Premise
2 | (B > (A & C)) Premise
|----------------------------
3 || ~((A v B) > C) Assumption
||---------------------------
4 || ((A v B) & ~C) 3 Implication, DeMorgan's, Double Negation
5 || (~(A > C) v ~(B > C)) 4 Dist, DeM, DN, Impl
6 ||| A Assumption
|||--------------------------
7 ||| (B & C) 1, 6 Conditional Elimination
8 ||| C 7 Conjunction Elimination
9 || (A > C) 6-8 Conditional Introduction
10 ||| B Assumption
|||--------------------------
11 ||| (A & C) 2, 10 Conditional Elimination
12 ||| C 11 Conjunction Elimination
13 || (B > C) 10-12 Conditional Introduction
14 || ~((A > C) & (B > C)) 5 DeMorgan's
15 || ((A > C) & (B > C)) 9, 13 Conjunction Introduction
16 | ((A v B) > C) 3-15 Negation Elimination (Reductio)
```

Last edited:

- #4

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- #5

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Code:

```
1| (A>(B&C)) {Premise}
2| (B>(A&C)) {Premise}
3|| ~((AvB)>C) {Assumption}
4|| (AvB) {from 3}
5|| ~C {from 3}
6||| A {Assumption}
7||| (B&C) {From 6 and 1}
8||| C {from 7}
9|| ~A {from 6; 8 contradicts 5}
10|| B {from 9 and 4}
11|| (A&C) {from 10 and 2}
12|| A {from 11}
13| ((AvB)>C) {from 3; 12 contradicts 9}
```

- #6

Science Advisor

Homework Helper

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1. Prove that:

[tex]\{\mathbf{[}([X \wedge Z] \wedge Y) \vee (\neg X \supset \neg Y)\mathbf{]},\, \mathbf{[}X \supset Z\mathbf{]},\, \mathbf{[}Z \supset Y\mathbf{]}\} \vdash \mathbf{[}X \equiv Y\mathbf{]}[/tex]

2. Show that the following is deductively inconsistent:

[tex]\{\mathbf{[}(\exists x)(\exists y)Fxy\, \vee \, (\forall x)(\forall y)(\forall z)Hxxyz\mathbf{]},\, \mathbf{[}(\exists x)(\exists y)Fxy \supset \neg Haaab\mathbf{]},\, \mathbf{[}(Hbbba\, \vee \, \neg Haaab)\equiv (\forall x)\neg (Ax\, \vee \, \neg Ax)\mathbf{]}\}[/tex]

3. Prove:

[tex]\{(\forall x)[(Fx\, \wedge \, \neg Kx) \supset (\exists y)([Fy\, \wedge \, Hyx]\, \wedge \, \neg Ky)],\, \mathbf{[}(\forall x)([Fx\, \wedge \, (\forall y)([Fy\, \wedge \, Hyx] \supset Ky)]\supset Kx)\supset M\mathbf{]}\} \vdash M[/tex]

I did what I could to ensure all the problems were legible, but ask for clarification if necessary.

- #7

Science Advisor

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1. Actually, this one is easy, I don't know why I put it. I can't think of a hint that doesn't give it away, so only readi this hint if you really need it. Prove that X > Y using hypothetical syllogism, prove that Y > X using disjunction elimination and the first premise.

2. Start by looking at the right half of the third sentence in the set.

3. Basically, you're proving that {

- #8

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Code:

```
1| (((X&Z)&Y)v(~X>~Y)) [premise]
2| (X>Z) [premise]
3| (Z>Y) [premise]
4|| X [assumption]
5|| Z [modus ponens; 4 and 2]
6|| Y [modus ponens; 5 and 3]
7| (X>Y) [conditional proof; 4 and 6]
8|| ~(~X>~Y) [assumption]
9|| ~X [forgot the name;8]
10|| Y [forgot;8]
11|| ((X&Z)&Y) [disjunctive syllogism;8 and 1]
12|| (X&Z) [simplification; 11]
13|| X [simplification; 12]
14| (~X>~Y) [reductio;8, 13 contradicts 9]
15| (Y>X) [transposition; 14]
16| ((X>Y)&(Y>X)) [conjunction; 15 and 7]
17| (X_=Y) [biconditional introduction; 16
```

Can anyone remind of the rule I forgot? I looked it up but couldn't find it.

- #9

Science Advisor

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Implication - ~(~X > ~Y) <> ~(~~X v ~Y)

DeMorgan's - ~(~~X v ~Y) <> ~~~X & ~~Y

Double Negation - ~~~X & ~~Y <> ~X & ~~Y

Simplification - ~X & ~~Y :> ~X

I don't know what you mean by using "relations" in 2 and 3. It's predicate logic.

- #10

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- #11

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Mx = x met John in New York

You could have a 2-place predicate N defined by:

Nxy = x met y in New York

A 3-place predicate P:

Pxyz = x met y in z

I don't know if the above explanation will help you though, because I can't imagine that you could know the rules of inference required to do the proof but not have learned about predicates of several terms.

The purpose of a "Logic Q&A Game" is to provide a fun and interactive way to learn and practice logical reasoning skills. It is designed to improve critical thinking and problem-solving abilities by presenting challenging arguments and asking players to prove their validity using the reductio ad absurdum method.

"Prove Argument w/ Reductio Ad Absurdum" is a logical reasoning method that involves disproving an argument by showing that its conclusion leads to an absurd or self-contradictory result. This method is commonly used in mathematics and philosophy to evaluate the validity of arguments.

In the game, players are presented with a series of arguments and are asked to prove their validity using the reductio ad absurdum method. They must carefully analyze the argument, identify any flaws or contradictions, and then provide a counterexample or absurd outcome that proves the argument to be false.

Playing a "Logic Q&A Game" can improve critical thinking skills, logical reasoning abilities, and problem-solving capabilities. It also helps to develop analytical thinking, attention to detail, and the ability to identify and evaluate arguments. Additionally, it can be a fun and engaging way to learn and practice these important skills.

Yes, the "Logic Q&A Game: Prove Argument w/ Reductio Ad Absurdum" can be played by people of all ages and skill levels. The game offers different levels of difficulty, allowing players to start with easier arguments and work their way up to more challenging ones. It is also suitable for both beginners and more experienced players who want to improve their logical reasoning skills.

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