Inconsistency between \int_compare and \fp_compare when dealing with logical operators

Question

I understand that the less automatic solution to compare multiple integer logical statements is by using \int_compare_p:nNn {...}{...}{...} along with \bool_if:nTF {condition(s)}{true}{false}. My question is why does \fp_compare:nTF {...}{...}{...} works fine with logical operators but \int_compare:nTF {...}{...}{...} yield an error even though all of the values used are integers?

\documentclass{article}

\ExplSyntaxOn
\NewDocumentCommand{\compare}{mmm}{
  \fp_compare:nTF {#1}{#2}{#3}
}

\NewDocumentCommand{\compareHardcode}{mm}{
  \bool_if:nTF {
    \int_compare_p:nNn {1} < {2} && \int_compare_p:nNn {2} < {3}
  }{#1}{#2}
}

\NewDocumentCommand{\compareError}{mmm}{
  %will not work using \int_compare:nTF 
  \int_compare:nTF {#1}{#2}{#3}
}
\ExplSyntaxOff
\begin{document}
  \compare{1 < 2 && 2 < 3}{yes}{no} % prints yes
  \compareHardcode{yes}{no}         % prints yes
  % Erroneous line commented:
  % \compareError{1 < 2 && 2 < 3}{yes}{no}
\end{document}

Answer 1

Something like 2<3 is a valid ⟨fp expr⟩ and \fp_eval:n { 2<3 } returns 1. You can also use logical operators. Examples:

\documentclass{article}

\begin{document}

\fpeval{1<2}

\fpeval{1<2 && 2<3}

\fpeval{1<2 || 3<2}

\fpeval{1>2 || 3<2}

\end{document}

This prints

1
1
1
0

So \fp_compare:nTF { 1<2 && 2<3 } { yes } { no } returns “yes”, because the evaluation is 1, which is nonzero.

To the contrary, 1<2 is not a valid ⟨int expr⟩. Moreover, you cannot use \int_compare:nTF { 1<2 && 2<3 } { yes } { no }, because the argument is invalid: the argument has to be in the form

⟨int expr⟩⟨relation⟩⟨int expr⟩⟨relation⟩…⟨int expr⟩⟨relation⟩⟨int expr⟩

and logical operators are not allowed.