# Return to Answer

 3 added 866 characters in body edited Feb 27 at 8:50 Mauro ALLEGRANZA 29.2k22 gold badges2020 silver badges6969 bronze badges 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 edited Jan 27 at 11:20 Mauro ALLEGRANZA 29.2k22 gold badges2020 silver badges6969 bronze badges 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. 1 answered Jan 27 at 8:45 Mauro ALLEGRANZA 29.2k22 gold badges2020 silver badges6969 bronze badges 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".