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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".


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.



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"].

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".


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.

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".


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.



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"].

2 added 423 characters in body
source | link

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".


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.

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".

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".


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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source | link

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".