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[PVS-Help] what to do when consequents are logically opposite
Hi, all,
First of all, thanks Professor Sam Owre for his quick and effective reply
of my last question and so I can continue my adventure in PVS :)
Right now I met a problem during a proof (PVS version is 3.1), the detail
is below:
[1] OO?(PROJ_1(x!1))
|---------
{1} FORALL (t: {x: T| x > ALPHA(x!1) AND x < OMEGA(x!1)}):
NOT tp1!1(t, x!1)
[2] FORALL (t: {x: T| x > ALPHA(x!1) AND x < OMEGA(x!1)}):
tp1!1(t, x!1)
where tp1!1 is a function returning boolean value. To me the consequences
1 and 2 are logically opposite (I have specified a lemma below for it and
checked the validated by PVS already):
Uni_n_Emp_temp: LEMMA
(forall (y: II): forall (x: {xx: T| x > ALPHA(y) and x < OMEGA(y)}): NOT tp1(x, y))
OR
(forall (y: II): forall (x: {xx: T| x > ALPHA(y) and x < OMEGA(y)}): tp1(x, y))
=> true;
Since in sequent calculus, consequences are disjunctive, right? and hence
the above proof should be correct in my opinion. Though PVS does not
behave the way :) Can anyone kindly enlighten me what to do in this
situation?
So I tried the command (merge-fnums (1, 2)), and thought that may solve
the problem by merging two consequences together. However, this time PVS
claims the following error message:
------------- start of the message ------------
Rule? (merge-fnums (1, 2))
No change on: (skip)
Uni_n_Emp.2.1.2 :
[1] OO?(PROJ_1(x!1))
|---------
{1} FORALL (t: {x: T| x > ALPHA(x!1) AND x < OMEGA(x!1)}):
NOT tp1!1(t, x!1)
[2] FORALL (t: {x: T| x > ALPHA(x!1) AND x < OMEGA(x!1)}):
tp1!1(t, x!1)
Rule?
Ill-formed rule/strategy: 2
No change on: (skip)
Error: Attempt to take the car of 2 which is not listp.
[condition type: simple-error]
Restart actions (select using :continue):
0: Return to Top Level (an "abort" restart).
1: Abort entirely from this process.
[1] pvs(41):
------------ end of the message ---------------
Did anyone encounter the similar problem before, and could you please
share your experience?
Thank you in advance.
Cheers
Chunqing