The transformations are as follows: (not P and not Q) or (not Q and not R)

valtricotinevh

valtricotinevh

Answered question

2022-07-16

I'm studying for my upcoming discrete math test and I'm having trouble understanding some equivalences I found in a book on the subject. I guess I'm not really familiar with these rules and I would like someone to walk me through the steps if they don't mind.
I know the elementary laws, De-Morgan's, absorption, distribution, associativity, symmetry, and idempotent laws. But I don't recognize how this person transforms the predicates. Could someone point out the name of the law I need to study?
The transformations are as follows:
( n o t   P   a n d   n o t   Q )   o r   ( n o t   Q   a n d   n o t   R )
( n o t   P   o r   ( n o t   Q   o r   n o t   R ) )   a n d   ( n o t   Q   o r   ( n o t   Q   o r   n o t   R ) )
( n o t   P   o r   n o t   Q   o r   n o t   R )   a n d   ( n o t   Q   o r   n o t   Q   o r   n o t   R )
( n o t   P   o r   n o t   Q   o r   n o t   R )   a n d   ( n o t   Q   o r   n o t   R )

Answer & Explanation

Selden1f

Selden1f

Beginner2022-07-17Added 14 answers

Step 1
The only step that should cause any trouble is the first, the equivalence of
( ¬ P ¬ Q ) ( ¬ Q ¬ R )
with ( ¬ P ( ¬ Q ¬ R ) ) ( ¬ Q ( ¬ Q ¬ R ) ) ;
from that to ( ¬ P ¬ Q ¬ R ) ( ¬ Q ¬ Q ¬ R )
is just removing redundant parentheses, and from that to ( ¬ P ¬ Q ¬ R ) ( ¬ Q ¬ R )
just uses the fact that X X is equivalent to X.
The first step is actually impossible as it stands: if Q is false and P and R are true, the second line is true, but the first is false. It appears from the bulk of the problem that the first line was supposed to read ( ¬ P ¬ Q ) ( ¬ Q ¬ R ) ;
if that is indeed the case, the first step is just an application of distributivity of ∨ over ∧, i.e., of the equivalence of ( X Y ) Z with ( X Z ) ( Y Z ); ¬ P is the X, ¬ Q is the Y, and ( ¬ Q ¬ R ) is the Z.
kadejoset

kadejoset

Beginner2022-07-18Added 2 answers

Step 1
(1) Put the argument in adequate symbolism:
Statements in ordinary language like the above reasoning may be misleading, ambiguous confusing. Therefore, let us prove it an adequate symbollism. Let '¬' stand for negation, '∧' for conjunction and '∨' for disjunction. Is the reasoning below what you mean?
(1) ( ¬ P ¬ Q ) ( ¬ Q ¬ R ) ( ¬ P ( ¬ Q ¬ R ) ) ( ¬ Q ( ¬ Q ¬ R ) ) (2) ( ¬ P ¬ Q ¬ R ) ( ¬ Q ¬ Q ¬ R ) (3) ( ¬ P ¬ Q ¬ R ) ( ¬ Q ¬ R )
One might say that the first step (1) is an application of distributivity of ∨ over ∧. I am afraid this would be incorrect.
As we know, the distributivity of ∨ over ∧, in symbols:
( ϕ ψ ) σ ( ϕ σ ) ( ψ σ )
should state that:
(1*) ( ¬ P ϕ ¬ Q ψ ) ( ¬ Q ¬ R ) σ ( ¬ P ϕ ( ¬ Q ¬ R ) σ ) ( ¬ Q ψ ( ¬ Q ¬ R ) σ )
Which as we see in the formalism above, is not the case. In fact (1) is not a logical equivalence at all (you can check it by a simple truth table).
(2) Typo:
We probably have a typo in the OP's question: presumably, it was supposed to be either
( n o t   P   a n d   n o t   Q )   o r   ( n o t   Q   a n d   n o t   R )
( n o t   P   o r   ( n o t   Q   a n d   n o t   R ) )   a n d   ( n o t   Q   o r   ( n o t   Q   a n d   n o t   R ) )
( n o t   P   o r   n o t   Q   a n d   n o t   R )   a n d   ( n o t   Q   o r   n o t   Q   a n d   n o t   R )
( n o t   P   o r   n o t   Q   a n d   n o t   R )   a n d   ( n o t   Q   a n d   n o t   R )
or ( n o t   P   a n d   n o t   Q )   o r   ( n o t   Q   o r   n o t   R )
( n o t   P   o r   ( n o t   Q   o r   n o t   R ) )   a n d   ( n o t   Q   o r   ( n o t   Q   o r   n o t   R ) )
( n o t   P   o r   n o t   Q   o r   n o t   R )   a n d   ( n o t   Q   o r   n o t   Q   o r   n o t   R )
( n o t   P   o r   n o t   Q   o r   n o t   R )   a n d   ( n o t   Q   o r   n o t   R )
Given the argument's structure, I bet in the second one (the first one is nonsense). In this case, we have the following correction of the formal reasoning above:
(1') ( ¬ P ¬ Q ) ( ¬ Q ¬ R ) ( ¬ P ( ¬ Q ¬ R ) ) ( ¬ Q ( ¬ Q ¬ R ) ) (2') ( ¬ P ¬ Q ¬ R ) ( ¬ Q ¬ Q ¬ R ) (3') ( ¬ P ¬ Q ¬ R ) ( ¬ Q ¬ R )
In (1') we have an application of the distributivity of ∨ over ∧.
In (2') we are justified by the associativity of ∨.
In (3') we have an application of the idempotency of ∨.

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