RISC JKU

Prove:

        (d)    [Graphics:Images/PND-nclever_gr_1.gif],

with no assumptions.

We prove (d) by the deduction rule.

We assume

        (1)    [Graphics:Images/PND-nclever_gr_2.gif]

and show

        (2)    [Graphics:Images/PND-nclever_gr_3.gif].

Formula (1) is transformed into:

        (3)    [Graphics:Images/PND-nclever_gr_4.gif].

We prove (2) by case distinction using (3).

Case  (4) [Graphics:Images/PND-nclever_gr_5.gif]:

Formula (4) is simplified to:

        (6)    [Graphics:Images/PND-nclever_gr_6.gif].

By (6) we can take appropriate values such that:

        (7)    [Graphics:Images/PND-nclever_gr_7.gif].

Because (7) matches a part of (2) , we can proceed in several ways.

Alternative proof 1: proved

For proving (2) it is sufficient to prove:

        (8)    [Graphics:Images/PND-nclever_gr_8.gif].

Formula (8) is true because it is subsumed by (7).

Alternative proof 2: pending

Pending proof of (2).

Case  (5) [Graphics:Images/PND-nclever_gr_9.gif]:

We prove (2) by contradiction.

We assume

        (9)    [Graphics:Images/PND-nclever_gr_10.gif],

and show [Graphics:Images/PND-nclever_gr_11.gif]

Formula (9) is simplified to:

        (10)    [Graphics:Images/PND-nclever_gr_12.gif].

Formula (10) is simplified to:

        (11)    [Graphics:Images/PND-nclever_gr_13.gif].

Formula (11) is simplified to:

        (12)    [Graphics:Images/PND-nclever_gr_14.gif].

The conjunction (12) splits into (13) , (14).

Formula (a contradiction) is proved because (14) and (5) are contradictory.


Converted by Mathematica      February 11, 2000