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[PVS-Help] PVS 4, errors in Proof buffer?
Does anyone recognize this symptom?
I'm running PVS 4 on some old sources and get an error in the proof-buffer.
The full buffer content is attached, but the last few lines are:
<PVS *proof* buffer>
.
.
.
Q.E.D.
Help! 12 nested errors. KERNEL:*MAXIMUM-ERROR-DEPTH* exceeded.
Debug (type H for help)
(UNIX::SIGSEGV-HANDLER #<unused-arg> #<unused-arg> #.(SYSTEM:INT-SAP
#x3FFFCA64))
Source:
*
I removed existing *.pvscontext* and *pvsbin/ *before invoking the Prover.
*M-x step-proof* does not generate this error.
Any ideas?
Thanks,
Steve Johnson
Starting pvs-cmulisp -qq ...
Warning: #<Command Line Switch "qq"> is an illegal switch
CMU Common Lisp 19d (19D), running on yosemitesam.cs.indiana.edu
With core: /nfs/nfs2/home/data/software/rhel/pvs4.1/bin/ix86-Linux/runtime/pvs-cmulisp.core
Dumped on: Tue, 2007-10-30 17:55:38-04:00 on photon.csl.sri.com
See <http://www.cons.org/cmucl/> for support information.
Loaded subsystems:
Python 1.1, target Intel x86
CLOS based on Gerd's PCL 2004/04/14 03:32:47
;;; Opening as shared library /l/pvs/bin/ix86-Linux/runtime/mu.so ...
;;; Done.
;;; Opening as shared library /l/pvs/bin/ix86-Linux/runtime/ws1s.so ...
;;; Done.
; Loading #P"/nfs/nfs2/home/data/software/rhel/pvs4.1/bin/ix86-Linux/runtime/bdd-cmu.x86f".
; Loading #P"/nfs/nfs2/home/data/software/rhel/pvs4.1/bin/ix86-Linux/runtime/mu-cmu.x86f".
; Loading #P"/nfs/nfs2/home/data/software/rhel/pvs4.1/bin/ix86-Linux/runtime/dfa-foreign-cmu.x86f".
*
*
A_0 :
|-------
{1} ((p => q) AND p) => q
Rule? (flatten)
Applying disjunctive simplification to flatten sequent,
this simplifies to:
A_0 :
{-1} (p => q)
{-2} p
|-------
{1} q
Rule? (split)
Splitting conjunctions,
this yields 2 subgoals:
A_0.1 :
{-1} q
[-2] p
|-------
[1] q
which is trivially true.
This completes the proof of A_0.1.
A_0.2 :
[-1] p
|-------
{1} p
[2] q
which is trivially true.
This completes the proof of A_0.2.
Q.E.D.
Run time = 0.05 secs.
Real time = 1.605 secs.
NIL
*
A_3 :
|-------
{1} NOT (p AND q) IFF (NOT p OR NOT q)
Rule? (split)
Splitting conjunctions,
this yields 2 subgoals:
A_3.1 :
|-------
{1} NOT (p AND q) IMPLIES (NOT p OR NOT q)
Rule? (flatten)
Applying disjunctive simplification to flatten sequent,
this simplifies to:
A_3.1 :
{-1} p
{-2} q
|-------
{1} (p AND q)
Rule? (split)
Splitting conjunctions,
this yields 2 subgoals:
A_3.1.1 :
[-1] p
[-2] q
|-------
{1} p
which is trivially true.
This completes the proof of A_3.1.1.
A_3.1.2 :
[-1] p
[-2] q
|-------
{1} q
which is trivially true.
This completes the proof of A_3.1.2.
This completes the proof of A_3.1.
A_3.2 :
|-------
{1} (NOT p OR NOT q) IMPLIES NOT (p AND q)
Rule? (flatten)
Applying disjunctive simplification to flatten sequent,
this simplifies to:
A_3.2 :
{-1} (NOT p OR NOT q)
{-2} p
{-3} q
|-------
Rule? (split)
Splitting conjunctions,
this yields 2 subgoals:
A_3.2.1 :
[-1] p
[-2] q
|-------
{1} p
which is trivially true.
This completes the proof of A_3.2.1.
A_3.2.2 :
[-1] p
[-2] q
|-------
{1} q
which is trivially true.
This completes the proof of A_3.2.2.
This completes the proof of A_3.2.
Q.E.D.
Help! 12 nested errors. KERNEL:*MAXIMUM-ERROR-DEPTH* exceeded.
Debug (type H for help)
(UNIX::SIGSEGV-HANDLER #<unused-arg> #<unused-arg> #.(SYSTEM:INT-SAP #x3FFFCA64))
Source:
*