请求证明指定逻辑表达式与s的等价性（禁用真值表）

Got it, let's break down this proof step by step—no truth tables needed, just standard logical equivalence rules. We'll start with your original expression ( s = P \land (R \to \lnot(Q \land P)) ) and transform it until we reach the target expression ( (P\land\lnot Q)\lor((P\land(\lnot R)) \land Q) ).

Step 1: Convert Implication to Disjunction #

First, recall that implication equivalence tells us ( A \to B \equiv \lnot A \lor B ). Apply this to ( R \to \lnot(Q \land P) ):
[
s = P \land (\lnot R \lor \lnot(Q \land P))
]

Step 2: Apply De Morgan's Law #

Use De Morgan's Law (( \lnot(A \land B) \equiv \lnot A \lor \lnot B )) on ( \lnot(Q \land P) ):
[
P \land (\lnot R \lor (\lnot Q \lor \lnot P))
]

Step 3: Remove Redundant Parentheses (Associative Law) #

Disjunction is associative, so ( (A \lor B) \lor C \equiv A \lor B \lor C ). We can simplify the inner expression:
[
P \land (\lnot R \lor \lnot Q \lor \lnot P)
]

Step 4: Distribute ( P ) Over the Disjunction #

Use the distributive law (( A \land (B \lor C \lor D) \equiv (A \land B) \lor (A \land C) \lor (A \land D) )) to expand the expression:
[
(P \land \lnot R) \lor (P \land \lnot Q) \lor (P \land \lnot P)
]

Step 5: Eliminate the Contradiction #

By the negation law, ( P \land \lnot P \equiv \text{False} ). Then, using the identity law for disjunction (( A \lor \text{False} \equiv A )), we can drop the contradiction term:
[
(P \land \lnot R) \lor (P \land \lnot Q)
]

Step 6: Split ( P \land \lnot R ) Using a Tautology #

To introduce ( Q ) into the first term (matching the target expression), we use the identity law with a tautology: ( A \equiv A \land (Q \lor \lnot Q) ) (since ( Q \lor \lnot Q \equiv \text{True} ), and ( A \land \text{True} \equiv A )):
[
[(P \land \lnot R) \land (Q \lor \lnot Q)] \lor (P \land \lnot Q)
]

Step 7: Distribute Again to Separate ( Q ) Terms #

Apply the distributive law to expand the first term:
[
[(P \land \lnot R \land Q) \lor (P \land \lnot R \land \lnot Q)] \lor (P \land \lnot Q)
]

Step 8: Combine Like Terms #

Use the associative law to group the last two terms, then reverse the distributive law to factor out ( P \land \lnot Q ):
[
(P \land \lnot R \land Q) \lor [(P \land \lnot Q) \land (\lnot R \lor \text{True})]
]
By the domination law, ( \lnot R \lor \text{True} \equiv \text{True} ), so this simplifies to:
[
(P \land \lnot R \land Q) \lor (P \land \lnot Q)
]

Step 9: Reorder Terms (Commutative Law) #

Finally, use the commutative law for disjunction (( A \lor B \equiv B \lor A )) to match the order of the target expression:
[
(P \land \lnot Q) \lor ((P \land \lnot R) \land Q)
]

This matches exactly the expression you needed to prove equivalent to ( s )! Every step uses a standard logical equivalence rule, so we've confirmed the equivalence without relying on truth tables.

内容的提问来源于stack exchange，提问作者Lugianite