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The resolvent of the set of clauses $(A \vee B, \sim A \vee D, C \vee \sim B)$ is 

  1. $A \vee​ B$
  2. $C \vee​ D$
  3. $A \vee​ C$
  4. $A \vee​ D$ 
in Mathematical Logic by Boss (30.2k points)
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3 Answers

+2 votes
Best answer

ans will be B) C   D

AB,AD,CB

A & ~A will cancel out  and so will B and ~B 

hence ans  

by Boss (48.8k points)
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0

Plz Tell the procedure of finding resolvent.

+1

Robinson in 1965 introduced the resolution principle, which can be directly 
applied to any set of clauses. The principal is 
"Given any two clauses A and B, if there is a literal P1 in A which has a 
complementary literal P2 in B, delete P1 & P2 from A and B and construct a 
disjunction of the remaining clauses. The clause so constructed is called 
resolvent of A and B." 
For example, consider the following clauses 
A: P V Q V R 
B: p' V Q V R 
C: Q' V R 
Clause A has the literal P which is complementary to `P in B. Hence both of 
them deleted and a resolvent (disjunction of A and B after the complementary 
clauses are removed) is generated. That resolvent has again a literal Q whose 
negation is available in C. Hence resolving those two, one has the final 
resolvent. 
A: P V Q V R (given in the problem) 
B: p' V Q V R (given in the problem) 
D: Q V R (resolvent of A and B) 
C: Q' V R (given in the problem) 
E: R (resolvent of C and D)

+1 vote

Resolvent is the outcome of applying the resolution rule.

$\frac{{A\lor B , \thicksim A \lor B, C \lor \thicksim D}}{C \lor D}$

Hence,Option(B) $C \lor D$.

by Boss (41k points)
0 votes

 

ans is B

by Active (1.9k points)

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