Question

Let the collection of nonlogical axioms be Σ ={(A(x)∧A(x))→B(x,y),A(x),B(x,y)→A(x)} and let the rule of inference be modus ponens,For each of the following, decide if it is a deduction.

If it is not a deduction, explain how you know that it is not a deduction.
(a)
A(x)
A(x)∧A(x)
(A(x)∧A(x))→B(x,y)
B(x,y)
​
(b)
B(x,y)→A(x)
A(x)
B(x,y)
​
(c)
(A(x)∧A(x))→B(x,y)
B(x,y)→A(x)
(A(x)∧A(x))→A(x)

Answer 1

The given sequences of deductions are not valid as they do not adhere to the structure required for modus ponens, where an antecedent must be affirmed to derive a conclusion.

Step-by-step explanation:

When evaluating deductions using a given set of axioms and the rule of inference modus ponens, we need to determine if each sequence leads from premises to a correct conclusion. Modus ponens allows us to derive a conclusion from a conditional statement and its antecedent.

For the given deductions:

• (a) Not a valid deduction because, while A(x) and A(x) ∧ A(x) are given, the latter does not serve as an antecedent to B(x,y).
• (b) Not a valid deduction because the sequence begins with the consequence B(x,y) → A(x) and attempts to affirm A(x), leading to B(x,y), which is not a form of modus ponens.
• (c) Not a valid deduction because it only presents conditionals without affirming any antecedent, thus no conclusion can be directly derived via modus ponens.

Each case lacks the necessary structure to guarantee the truth of the conclusion based on the premises. Deductive reasoning requires a valid argument structure to ensure that if the premises are true, the conclusion is also true.

Answer 2

(a) A(x), A(x)∧A(x), (A(x)∧A(x))→B(x,y), B(x,y) is a deduction.

(b) B(x,y)→A(x), A(x), B(x,y) is not a deduction.

(c) (A(x)∧A(x))→B(x,y), B(x,y)→A(x), (A(x)∧A(x))→A(x) is a deduction.

Step-by-step explanation:

To determine if each of the given statements is a deduction, we need to apply the rule of inference, modus ponens, to the nonlogical axioms in the collection Σ.

(a) A(x), A(x)∧A(x), (A(x)∧A(x))→B(x,y), B(x,y)

To check if this is a deduction, we need to see if we can derive B(x,y) from the given statements using modus ponens. Let's go step by step:

1. From A(x) and (A(x)∧A(x))→B(x,y), we can apply modus ponens to conclude B(x,y).

2. Since B(x,y) is one of the given statements, we have successfully derived it.

Therefore, the given statements (a) A(x), A(x)∧A(x), (A(x)∧A(x))→B(x,y), B(x,y) form a deduction.

(b) B(x,y)→A(x), A(x), B(x,y)

To check if this is a deduction, we need to see if we can derive A(x) from the given statements using modus ponens. Let's go step by step:

1. From B(x,y)→A(x) and B(x,y), we cannot directly apply modus ponens because the order of the statements is reversed. Modus ponens allows us to infer the consequent (A(x)) from the antecedent (B(x,y)).

2. Therefore, we cannot derive A(x) from the given statements.

Hence, the given statements (b) B(x,y)→A(x), A(x), B(x,y) are not a deduction.

(c) (A(x)∧A(x))→B(x,y), B(x,y)→A(x), (A(x)∧A(x))→A(x)

To check if this is a deduction, we need to see if we can derive A(x) from the given statements using modus ponens. Let's go step by step:

1. From (A(x)∧A(x))→B(x,y) and (A(x)∧A(x))→A(x), we can apply modus ponens to conclude B(x,y).

2. From B(x,y)→A(x) and B(x,y), we can apply modus ponens to conclude A(x).

3. Since A(x) is one of the given statements, we have successfully derived it.

Therefore, the given statements (c) (A(x)∧A(x))→B(x,y), B(x,y)→A(x), (A(x)∧A(x))→A(x) form a deduction.