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I have hard time determining whether the below argument is valid:

(P1) Nobody ever plays with a bear.

(P2) Bob plays with a bear.

(C) Therefore, bears are friendly.

I think this argument is valid. Suppose this argument is invalid. Then the premises P1 and P2 are true while the conclusion C is false. We get a contradiction if both P1 and P2 are jointly true. So, the argument must be valid, by the law of excluded middle.

Is my reasoning correct?

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  • 3
    This is why contradictions cause logical explosions, AKA trivialization, because they validate whichever conclusion one attaches to them. So in one sense you're right (the argument is technically, very very technically, valid) but in another wrong (the argument is meta-invalid, so to speak). Oct 8, 2023 at 0:17
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    It is valid, but not by the law of excluded middle. It is valid by the law of explosion: contradiction implies anything. This is because it leaves no room for counterexamples - no object can satisfy contradictory premises to become one.
    – Conifold
    Oct 8, 2023 at 1:13
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    Bob may be a AI bot properly named after its creator which is not remotely unlikely. Thus if your first premise 'Nobody' only ranges over humans as colloquially understood, then there's no contradiction or explosion from your two premises and then the argument sounds possibly cogent. Bears are friendly (to bots)... Oct 8, 2023 at 4:38
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    This question has come up several times before, e.g. here: philosophy.stackexchange.com/questions/84490/… It is called the principle of explosion and is valid in classical logic, though not all logics.
    – Bumble
    Oct 8, 2023 at 4:53
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    Does this answer your question? How can syllogisms with contradictory premises be valid? Oct 8, 2023 at 11:07

3 Answers 3

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This is valid in classical logic because of the principle of explosion.

P1: ¬∃x∃y (B(x) ∧ P(y,x))
P2: ∃x (B(x) ∧ P(b,x)
C: ∀x (B(x) → F(x))
B(x) is "x is a bear", and P(x,y) is "x plays with y", and b is Bob. C is unimportant.

 1. P1
 2. P2
 3. | B(a) ∧ P(b, a)            Assumption for existential elim
 4. | ∃y (B(a) ∧ P(y, a))       Existential Intro, 3
 5. | ∃x∃y (B(x) ∧ P(y, x))     Existential Intro, 4
 6. ∃x∃y (B(x) ∧ P(y, x))       Existential Elim, 3-5
 7. ∃x∃y (B(x) ∧ P(y, x)) ∨ C   Or Intro
 8. C                           Disjunctive Syllogism, 1,7 
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  • I'm not sure why you are treating 3 as an assumption. It follows from P2. You could skip lines 3 to 5 and just say that 6 follows from P2 by existential generalization.
    – Bumble
    Mar 8 at 14:10
  • I can’t skip lines 3 to 5 because Existential intro will put the exists y in front of exists x. Line 3 is an assumption because there’s the requirement that “a” is an eigenvariable, so by treating it as an assumption, we can easily check that in line 5, the eigenvariable is gone, and so we can safely discharge the assumption Mar 8 at 15:42
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Is this [a] valid argument?

This line of reasoning is too difficult to analyze. First, the two premises contradict each other. The first is an "O" statement: No person is someone who plays with bears.

The second is an "I" statement: Some people (Bob) are persons who play with bears. O and I statements are mutually contradictory.

Second, the apparent middle term ("bears") is still present in the conclusion. The middle term connects the two premises, and disappears from the conclusion.

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V = The OP's argument is valid.
P = Nobody plays with bears
~P = Bob (somebody) plays with bears
C = Bears are friendly

The OP's argument about his own argument:

  1. V or ~V
  2. ~V [Assume for reductio ad absurdum]
  3. If ~V then ((P & ~P) & C) [premise]
  4. (P & ~P) & C [2, 3 Modus Ponens]
  5. P & ~P [4 Simplification]
  6. ~~V [2 to 5 reductio ad absurdum]
  7. V [1, 6 Disjunctive Syllogism/LEM]

QED.

How does C follow from P & ~P?

Ex Contradictione Sequitur Quodlibet

  1. P & ~P [premise]
  2. P [1 Simplification]
  3. P v C [2 Addition]
  4. ~P [1 Simplification]
  5. C [3, 4 Disjunctive Syllogism]
    QED

I don't see any errors. Does anyone else?

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