• Banno
    14.3k
    Vann McGee claims that modus ponens "is not strictly valid" in an article from 1985

    Opinion polls taken just before the1980 election showed the Republican Ronald Reagan decisively ahead of the Democrat Jimmy Carter, with the other Republican in the race, John Anderson, a distant third. Those apprised of the poll results believed, with good reason:
    [1] If a Republican wins the election, then if it's not Reagan who wins it will be Anderson.
    [2] A Republican will win the election. Yet they did not have reason to believe
    [3] If it's not Reagan who wins, it will be Anderson

    Curious.

    I only have this example. Does anyone have more?
  • TheMadFool
    12k
    The conditional [1] is false.
  • Michael
    9.8k
    A → (B → C) is equivalent to (A ∧ B) → C (see Exportation (logic)). So the argument above is:

    1. (A ∧ B) → C
    2. A
    3. B → C

    Which isn't modus ponens.
  • Banno
    14.3k
    Indeed. That' the point.
  • TheMadFool
    12k
    How?Banno

    [1] If a Republican wins the election, then if it's not Reagan who wins it will be Anderson.

    The conditional is formulated as if the following is true,

    Republicans win then the vote ranking will be:

    1. Reagan
    2. Anderson
    3. Jimmy Carter

    It fails to consider the possibility that actually did occur:

    Republicans win then the vote ranking will be:

    1. Reagan
    2. Jimmy Carter
    3. Anderson
  • Banno
    14.3k
    Sure.

    But (1) is true.
  • Michael
    9.8k
    Indeed. That' the point.Banno

    I thought your point was that it was supposed to be an example of an invalid modus ponens? Except it's not a modus ponens at all.
  • Banno
    14.3k


    Modus ponens:
    If p then q
    p
    therefore, q

    p: A Republican wins the election,
    q: If it's not Reagan who wins, it will be Anderson

    So:

    If A Republican wins the election, then If it's not Reagan who wins, it will be Anderson

    and

    A Republican wins the election

    which, by MP, gives

    If it's not Reagan who wins, it will be Anderson.

    But if it is not Reagan who wins, it will be Jimmy Carter. So there is a prima facie case that MP reaches a false conclusion from true premises.
  • Michael
    9.8k
    See here. When you replace A → (B → C) with the logically equivalent (A ∧ B) → C you'll see that it isn't modus ponens. The former just gives the false appearance that it is one.
  • Banno
    14.3k


    Sure there are other ways to pars it.

    Unless you can rule out the MP parsing, showing that there are other parsings is irrelevant.

    Can you rule out the MP parsing?
  • Michael
    9.8k
    1. A → (B → C)
    2. ¬(B → C)
    3. ¬A

    If a Republican wins the election then if it's not Reagan who wins it will be Anderson
    If it's not Reagan who wins it won't be Anderson
    Therefore a Republican won't win the election

    Would you say that's an invalid modus tollens? I'd say it's not modus tollens. The argument is:

    1. (A ∧ B) → C
    2. ¬(B → C)
    3. ¬A
  • bongo fury
    1.2k
    YetBanno

    Weasel. Yes they did. If they "believed, with good reason" both [1] and [2], then they had deductive reason to believe [3].

    Deductive not good enough? Sure. Deductive not always good enough. Too strict at times. Then go inductive.

    "Strictly", though... deductively... modus ponens valid. Here, as anywhere.
  • Moliere
    1.8k
    Umm... I mean, if it's logically equivalent, then they are the same?

    Just as you can convert "A → (B → C)" to "(A ∧ B) → C", you can also convert it back.

    So " If a Republican wins the election, then if it's not Reagan who wins it will be Anderson" is logically equivalent, has the same truth-value, as "If a Republican wins the election And it is not Reagan Then it will be Anderson" (since "not" is being parsed as part of the sentence, and not an operator in the above form)


    Once you convert it you have to also convert premise 2 so that the antecedent includes two conditions.







    At first blush doesn't it seem like "B -> C" is false, though? Since clearly if Reagan does not win then it will be Carter. What am I missing?
  • Cuthbert
    360
    I think voters have reason to believe [3] if they rely solely on the two premisses [1] and [2]. But these two premisses do not contain all the relevant background information. Consider:

    [1] ... as above
    [2]..... as above
    [3] If it's not Reagan who wins, it will be Anderson
    [4] Anderson will not win [additional background information from the scenario]
    [5] Reagan will win [by modus tollens [3] and [4]]

    The trick is that [1] and [2] are selected parts of information. Crucially, [4] is not mentioned. If we did not know from the scenario that Anderson was a hopeless case, we would be quite happy to accept [3] following from [1] and [2] as indeed (I submit) it does follow - by modus ponens. So modus ponens is not threatened as a logical form. Phew.
  • Cuthbert
    360
    So if I'm right then it's a case of a premiss being smuggled out of an argument rather than smuggled in. Very neat.
  • Amalac
    409


    p: a Republican wins the election.
    q: if it's not Reagan who wins, it will be Anderson (q = r→s)

    r→s is equivalent to ¬r v s, so q should be interpreted as: either Reagan wins or Anderson wins.

    Since Reagan won, q is true, since one of the components of the disjunction is true.

    Before Reagan won, if it was possible that Carter could have won, they couldn't have known whether q is true or not.

    But, since p is true , q is true, and p→q is true, modus ponens leads to a true conclusion from true premises.

    It's just a case of the so called “paradoxes of material implication”.
  • TheMadFool
    12k
    Sure.

    But (1) is true.
    Banno

    P = Republicans won
    R = Reagan won
    A = Anderson won

    P -> (~R -> A)

    Let's ignore "P ->" for the moment.

    Is (~R -> A) true? No!

    The conditional as a whole, P -> (~R -> A) is false if P is true.
  • Bartricks
    3.9k
    1. If p, then premise 2 is false.
    2. P
    3. Therefore premise 2 is false.
  • Amalac
    409


    Premise 2 is false= ¬p

    1. p→¬p means: ¬p v ¬p, which is ¬p.
    2. p
    3. Therefore ¬p.

    Obviously, if you have both p and ¬p as premises you can conclude anything you want, since anything follows from a contradiction.

    The implication p→¬p is only true if ¬p is true. So if 2 is true, 1 is false, and you can't infer ¬p using modus ponens.

    See:

    It's just a case of the so called “paradoxes of material implication”.Amalac
  • sime
    611
    In the case of statistics or beliefs which involve probabilities,the standard non-probabilistic version of Modus Ponens is generally inapplicable,since there it isn't generally used as a constructive principle, and so it is neither fair nor surprising to point out the failure of MP in this situation . And yet statistical relations do obey a generalised version of Modus-Ponens with respect to conditional probabilities:

    Take for instance, the following beliefs:

    P (Reagan wins) = 0.80
    P (Carter wins ) = 0.15
    P (Andy wins ) = 0.05 (i.e. distant third republican)


    P (Reagan or Andy) = 0.80 + 0.05 = 0.85 (i.e. the probability that a Republican wins)

    P(Reagan | Reagan or Andy ) + P(Andy | Reagan or Andy) = 1 (i.e, as a logical tautology, Andy must win if Reagan doesn't, relative to the assumption that a republican wins)

    But if Reagan doesn't win, then

    P(Andy | Andy or Carter) = 0.05/ (0.05 + 0.15) = 0.25, (i.e. Carter remains favourite over Andy)

    But notice that although this example contradicts (the misuse of) logical Modus Ponens, it doesn't contradict "probabilistic modus ponens" of the form P (B,A) = P( B | A) * P(A), which when summed over the values permitted for A recovers P(B).

    In other words, if we take the conditional probabilities as being fundamental and follow this example in the bottom-up direction using this probabilistic modus-ponens, we recover the initial unconditional beliefs.
  • Bartricks
    3.9k
    1.if p, then modus ponens is invalid.
    2.p
    3. Therefore modus ponens is invalid.
  • TonesInDeepFreeze
    873
    'R' for 'Reagan wins'
    'A' for 'Anderson wins'
    'C' for 'Carter wins'
    'R v A' for 'a Republican wins'

    (R v A) -> (~R -> A)
    R v A
    therefore ~R -> A

    That's an instance of modus ponens..
  • fdrake
    4.9k
    Keep track of domains.

    [1] If a Republican wins the election, then if it's not Reagan who wins it will be Anderson.


    (A) If republican (x) and wins (x), then (B) (not (Reagan(x) implies Anderson (x).
    (A=>B) has people in general talked about in (A), contextually only republicans are talked about in (B).

    [2]A Republican will win the election.

    Has a different domain - it's talking about the set of presidential candidates - a republican - a republican candidate.

    If it's not Reagan who wins, it will be Anderson

    Is false in the overall domain {Anderson, Reagan, Carter}, but true in the domain {Anderson, Reagan}.

    So while you can string match "A Republican will win the election" with making A true, it doesn't follow that string matching preserves truth condition when the strings have different implicatures . Specifically, when you move from "If" to "then" in (A), it's coming along with a domain change (from candidates to republicans).

    "A republican will win the election"

    Has an implicit domain of presidential candidates.

    [1] If a Republican wins the election, then if it's not Reagan who wins it will be Anderson.

    then if it's not Reagan who wins it will be Anderson.

    Has an implicit domain of republicans.
  • TonesInDeepFreeze
    873
    Since R is the case, ~R -> A is true.

    Bu the puzzle includes an intensional operator "believe'.
  • TonesInDeepFreeze
    873
    Keep track of domains.fdrake

    We may state these atomic propositions purely as sentence letters so there is not a need to involve domains.
  • fdrake
    4.9k
    The propositions don't involve quantifiers. There's no issue of domains.TonesInDeepFreeze

    There's a set of candidates C:

    {Carter, Reagan, Anderson}

    [1] If a Republican wins the election, then if it's not Reagan who wins it will be Anderson.

    Maps {Carter, Reagan, Anderson} to {Reagan, Anderson}, the latter set of candidates is where the disjunction is understood and evaluates to true.

    [2] A Republican will win the election.

    Asserts that of {Carter, Reagan, Anderson}, the winner will be a member of {Reagan, Anderson}

    [3] If it's not Reagan who wins, it will be Anderson

    Asserts that of {Reagan, Anderson}, if Reagan doesn't win, it will be Anderson.

    There are domains in the sense that the disjunctions are implicitly quantifying over those sets. That is what I meant - in the same sense that existential quantification is iterated disjunction on a finite domain.
  • TonesInDeepFreeze
    873
    Premise 2 is falseAmalac

    At a point before the election, with 'wins' understood as 'will win', then R v A is true.

    At a point after the election, with 'wins' understood as 'won', then R v A is true.
  • Amalac
    409


    At a point before the election, with 'win's understood as 'will win', then R v A is true.

    At a point after the election, with 'wins' understood as 'won', then R v A is true.
    TonesInDeepFreeze

    I was refering to Bartricks' fallacious argument, not to the OP's.
  • TonesInDeepFreeze
    873


    Yes, but the puzzle (if there truly is one) maintains with or without resort to talking about domains.

    Include the premises:

    R
    ~C
    ~A

    Then

    (R v A) -> (~R -> A)
    R v A
    ~A
    ~C
    therefore ~R -> A

    Still valid.
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