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The Game of Modern Logic By Layman E. Allen Professor of Law and Senior Research Scientist, University of Michigan |
The Tarski Short Cut
(for testing whether a candidate
WFF is a provable theorem)
| Suppose the candidate to be false, and see whether that supposition leads to contradiction. If it does, the candidate is a provable theorem. In effect, this is using the No Rule (indirect proof). |
Truth Table Formulation of Two-Valued Propositional Logic
|
|
| p q | Kpq | Apq | Np | Cpq | Epq |
| 1 1 | 1 | 1 | 0 | 1 | 1 |
| 1 0 | 0 | 1 | 0 | 0 | 0 |
| 0 1 | 0 | 1 | 1 | 1 | 0 |
| 0 0 | 0 | 0 | 1 | 1 | 1 |
Three Different Decision Procedures
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|
| p q | Cpq | Nq | Np | CNqNp | C-Cpq-CNqNp |
| 1 1 | 111 | 01 | 01 | 10101 | 1-111-10101 |
| 1 0 | 010 | 10 | 01 | 01001 | 1-010-01001 |
| 0 1 | 101 | 01 | 10 | 10110 | 1-101-10110 |
| 0 0 | 100 | 10 | 10 | 11010 | 1-100-11010 |
| Since the candidate (Column 6) is true for all combinations of p and q, it is a provable theorem. |
|
|
| C-Cpq-CNqNp |
| 1-111-10101 |
| 1-010-01001 |
| 1-101-10110 |
| 1-100-11010 |
| Since the candidate is true for all combinations of p and q, it is a provable theorem. |
|
|
Example 1: C-Cpq-CNqNp
| Candidate | Comments |
| C-Cpq-CNqNp
0 110 01001 0 |
Suppose the candidate to be false, and see if that leads to contradiction by the definitions of its operators. It does: Cpq=1 and Cpq=0. The candidate is a theorem. C-Cpq-CNqNp = 1. |
| 1
267 34756
8 |
The numbers to the left indicate the order in which the truth values, 0 and 1, are assigned to the letters in the candidate expression by the supposition and the definitions of the operators. |
Example 2: C-NCpq-KpNq
| Candidate | Comments |
| C-NCpq-KpNq
0 1010 0110 1 |
By supposing C-Cpq-CNq-Np = 0, KpNq=0 and KpNq=1; therefore C-Cpq-CNq-Np = 1. |
| 1
2456 3576
8 |
The numbers to the left indicate the order in which the truth values, 0 and 1, are assigned. |
Example 3: E-NApq-KNpNq
| Candidate | Comments |
| E-NApq-KNpNq
0 1 0 Case 1 000 11010 1 2456 37586 9 0 1 Case 2 1000 1010 1 2867 34657 9 |
E-NApq-KNpNq
can be false in two different ways: (1) NApq=1 and KNpNq=0, and (2)
NApq=0 and KNpNq=1. Thus, two cases must be considered, and both must be
contradictory.
In Case 1 supposing E-NApq-KNpNq=0 leads to KNpNq=0 and KNpNq=1, and in Case 2 supposing it leads to NApq=0 and NApq=1. Thus, E-NApq-KNpNq=1. |
Example 4: C-CKpqr-ACprCqr
| Candidate | Comments |
| C-CKpqr-ACprCqr
0 11111 0010010 1 |
Supposing C-CKpqr-ACprCqr=0 leads to r=0 and r=1. Thus, C-CKpqr-ACprCqr=1. |
| 1
29687 3467587
10 10 |
Order of assignment of truth values. |
Example 5: C-CNpq-Apq
| Candidate | Comments |
| C-CNpq-Apq
0 1100 000 0 |
Supposing C-CNpq-Apq=0 leads to CNpq=1 and CNpq=1. Thus, C-CNpq-Apq=1. |
The Tarski Short Cut and Derivable Rules
| Testing Derivability of a Rule Candidate
Not only is the Tarski Short Cut a powerful tool for
ascertaining whether or not a WFF is a theorem, but it can also be used
to determine whether a rule candidate is a derivable rule. With every rule
candidate there is an associated C-WFF. If its associated C-WFF is a theorm,
then the rule candidate is derivable, and vice versa, that is:
Rule Candidate's Associated C-WFF The C-WFF associated with a rule candidate is the C-WFF such that its antecedent is the conjunction of the rule candidate's premisses and its consequent is the rule candidate's conclusion. Thus, the associated C-WFF for the rule candidate:
When the Tarski Short Cut is used to test C-KpArs-KrAsp,
it turns out not to be a theorem; so, the rule candidate is not derivable.
On the other hand, when the associated C-WFF to the
rule candidate:
And WFF 'N PROOF players should be able to construct a proof of this new derived rule, that is, that p,q,CArqs/Ai,Co,Ki is a Solution for the Goal, KAsrp. |
Advice to Coaches and Players of Adventurous WFF 'N PROOF
| Thorough familiarity with the Tarski Short Cut contributes remarkably to the quality of strategy that players are equipped to bring to bear upon their play in pursuit of the Leslie Nielsen Scholarship award. --LEA |
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