[OpenIndiana-discuss] PATH=/usr/xpg4/bin /bin/ksh -c "rm -f"

[James]

Hello, please can someone help me understand this xpg4, ksh, rm oddity? 
Normally: rm with no args complains and adding -f makes it accept no args.

$ rm
usage: rm [-fiRr] file ...
$ rm -f
$

On OpenSolaris [but not 10, 11.3] with /usr/xpg4/bin in the path it 
behaves differently but only with /bin/ksh or /bin/sh (ksh93)

$ PATH=/usr/xpg4/bin /bin/ksh -c "rm -f"
Usage: rm [-cFdfirRuv] file ...
$ PATH=/usr/xpg4/bin /bin/sh -c "rm -f"
Usage: rm [-cFdfirRuv] file ...
$ PATH=/usr/xpg4/bin /bin/zsh -c "rm -f"
$ PATH=/usr/xpg4/bin /bin/bash -c "rm -f"

Without xpg4 in the path:
$ PATH=/usr/bin /bin/ksh -c "rm -f"

Calling rm directly does what I expect:

$ PATH=/usr/xpg4/bin /bin/ksh -c "/usr/xpg4/bin/rm -f"
$ PATH=/usr/xpg4/bin /bin/ksh -c "/bin/rm -f"
$ PATH="" /bin/ksh -c "PATH=/usr/xpg4/bin rm -f"

$ PATH=/usr/xpg4/bin /bin/ksh -c "alias rm"
rm: alias not found

$ PATH=/usr/xpg4/bin /bin/truss -f /bin/ksh -c "rm -f" |& grep rm
4436:	lstat("/usr/xpg4/bin/rm", 0xFFFFFC7FFFDFDCE0)	= 0
Usage: rm [-cFdfirRuv] file ...
$ PATH= /bin/truss -f /bin/ksh -c "PATH=/usr/xpg4/bin rm -f" |& grep rm
4439:	execve("/usr/xpg4/bin/rm", 0x00AA80E0, 0x00AA8108)  argc = 2
4439:	resolvepath("/usr/xpg4/bin/rm", "/usr/xpg4/bin/rm", 1023) = 16
4439:	stat64("/usr/xpg4/bin/rm", 0x0803E900)		= 0

I can't see a built-in rm or alias in ksh and if it were I would expect 
-f to have the same effect.  The path must be making ksh act 
differently.  What am I not understanding here?