Illegal Herbrand Logic sentence

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I am studying the Stanford Introduction to Logic course. There was a problem about whether an expression is legal sentence of Herbrand Logic or not. It asks:




Say whether $p(f(p(a))$ is a syntactically legal sentence of Herbrand Logic. Assume that $a$ and $b$ are object constants, $f$ is a unary function constant, and $p$ is a unary relation constant.




The answer was 'illegal', but I don't understand why it's illegal. I suspect $p(a)$ might not be a term so that it cannot be applied to $f$, but I'm not sure. Why is the expression illegal?



(link to the problem: http://intrologic.stanford.edu/notes/chapter_09.html)




logic first-order-logic




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  • 2




    Your suspicion is almost correct; $p(a)$ is not a term, so $f$ cannot be applied to it.
    – Andreas Blass
    Jul 24 at 13:08










  • @AndreasBlass Thank you!!
    – JSong
    Jul 25 at 2:02














up vote
2
down vote

favorite












I am studying the Stanford Introduction to Logic course. There was a problem about whether an expression is legal sentence of Herbrand Logic or not. It asks:




Say whether $p(f(p(a))$ is a syntactically legal sentence of Herbrand Logic. Assume that $a$ and $b$ are object constants, $f$ is a unary function constant, and $p$ is a unary relation constant.




The answer was 'illegal', but I don't understand why it's illegal. I suspect $p(a)$ might not be a term so that it cannot be applied to $f$, but I'm not sure. Why is the expression illegal?



(link to the problem: http://intrologic.stanford.edu/notes/chapter_09.html)




logic first-order-logic




share|cite|improve this question

















  • 2




    Your suspicion is almost correct; $p(a)$ is not a term, so $f$ cannot be applied to it.
    – Andreas Blass
    Jul 24 at 13:08










  • @AndreasBlass Thank you!!
    – JSong
    Jul 25 at 2:02












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I am studying the Stanford Introduction to Logic course. There was a problem about whether an expression is legal sentence of Herbrand Logic or not. It asks:




Say whether $p(f(p(a))$ is a syntactically legal sentence of Herbrand Logic. Assume that $a$ and $b$ are object constants, $f$ is a unary function constant, and $p$ is a unary relation constant.




The answer was 'illegal', but I don't understand why it's illegal. I suspect $p(a)$ might not be a term so that it cannot be applied to $f$, but I'm not sure. Why is the expression illegal?



(link to the problem: http://intrologic.stanford.edu/notes/chapter_09.html)




logic first-order-logic




share|cite|improve this question













I am studying the Stanford Introduction to Logic course. There was a problem about whether an expression is legal sentence of Herbrand Logic or not. It asks:




Say whether $p(f(p(a))$ is a syntactically legal sentence of Herbrand Logic. Assume that $a$ and $b$ are object constants, $f$ is a unary function constant, and $p$ is a unary relation constant.




The answer was 'illegal', but I don't understand why it's illegal. I suspect $p(a)$ might not be a term so that it cannot be applied to $f$, but I'm not sure. Why is the expression illegal?



(link to the problem: http://intrologic.stanford.edu/notes/chapter_09.html)





logic first-order-logic





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 24 at 15:39









Taroccoesbrocco

3,45441431




3,45441431









asked Jul 24 at 10:20









JSong

1283




1283







  • 2




    Your suspicion is almost correct; $p(a)$ is not a term, so $f$ cannot be applied to it.
    – Andreas Blass
    Jul 24 at 13:08










  • @AndreasBlass Thank you!!
    – JSong
    Jul 25 at 2:02












  • 2




    Your suspicion is almost correct; $p(a)$ is not a term, so $f$ cannot be applied to it.
    – Andreas Blass
    Jul 24 at 13:08










  • @AndreasBlass Thank you!!
    – JSong
    Jul 25 at 2:02







2




2




Your suspicion is almost correct; $p(a)$ is not a term, so $f$ cannot be applied to it.
– Andreas Blass
Jul 24 at 13:08




Your suspicion is almost correct; $p(a)$ is not a term, so $f$ cannot be applied to it.
– Andreas Blass
Jul 24 at 13:08












@AndreasBlass Thank you!!
– JSong
Jul 25 at 2:02




@AndreasBlass Thank you!!
– JSong
Jul 25 at 2:02















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