Modus Tollens / Popperian Syllogism for the Black Swan

Question

If it is a swan, then it will always be white.

It seems like Popper is not asking so much people to build syllogisms, but rather DEPENDING on Modus Tollens as an engine for falsification.

So that, if I propose -

• If it is a swan, then it will always be white.

And then go ahead and execute my testing model by traveling to Australia.

Now I have:

If it is a swan, then it will always be white.

I just saw a black swan.

• If A then B
• Not B
• Therefore, Not A

But, here is my confusion: it seems like my hypothesis has some structural issue for Modus Tollens to take effect.

Using what I proposed I get:

• If it is a swan, then it will always be white.
• I just saw a black swan. [Not B]
• It is not a swan. [Not A]

The "it" here suffers dually - from being ambiguous, as you need to stop for a second and figure out that the "it" here refers to the Antecedent "it" in the Conditional statement.

But, WORSE! - if I were truly going to go with Not A, saying

"It is not a swan." [Not A]

makes no sense.

So, my question is: how should I structure the white swan hypothesis Conditional Statement so that the "Not A" side of the syllogism under the process of Modus Tollens looks a bit more trenchant?

Answer 1

It is quite simple.

The "general law" we have assumed as an hypothesis is :

"Every swan is white"

that, according to the language of predicate logic, is :

(1) "for every x, if x is a Swan, then x is White" [in symbols : ∀x(S(x) → W(x))].

Yoy are travelling in Australia and you find a black swan, that is :

(2) "there is an x such that, x is a Swan and x not is White" [in symbols : ∃x(S(x) ∧ ¬W(x))] .

Call that Swan s; from (2), by Existential instantiation we have :

S(s) ∧ ¬ W(s).

Form (1), by Universal instantiation we have :

S(s) → W(s).

But the two formula are contradictory, because P ∧ ¬Q is the negation of P → Q, and thus we have to conclude with the negation of the initial hypothesis, i.e. we have falsified the purported "general law" :

"Every swan is white".

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The argument above can be expressed also with Modus Tollens.

We have S(s) → W(s) and from S(s) ∧ ¬ W(s), by Simplification we get : S(s) and ¬ W(s).

Now we have : S(s) → W(s) and ¬ W(s) and by MT we conclude with :

¬ S(s).

This contradicts S(s) and we are done.

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It is not exactly a Syllogism because a syllogism needs three terms, like e.g.: S,P and M.

In our example, we have only two : Swan and White.

Thus the inference is :

1st premise : "All Swans are White" [A-type : Universal Affirmative : "All S are W"]

2nd premise : "Some Swans are not White" [O-type : Particular Negative : "Some S are not W"]

With them, we may have a valid Baroco syllogism, concluding in O-type : Particular Negative.

In fact, the conclusion we have reached denying the first premise :

"Not every Swan is White" ["Not all S are W"]

is a Particular Negative, because it amounts to "Some Swans are not White" ["Some S are not W"].