Discrete Mathematics -Propositional Logic

Question

How to get

B↔C from B∧C . Using derivation method i.e Step by Step .


My attempt :
            i) B∧C Rule P 
            ii)B Rule T (i) Simplification B∧C ⇒ B
            iii)C Rule T (i) Simplification B∧C ⇒ C

I'm stuck here ..

Answer 1

HINT (first 3 lines):

1. Suppose $B\land C$

2. Suppose $B$

3. $C$ (from 1)

Answer 2

I use Lukasiewicz/Polish notation and condensed detachment. The axioms I will use are:

1. CpCqp.
2. CKpqp.
3. CKpqq.
4. CCpqCCqpEpq.

Here we go:

assumption 5 Kbc.
D2.5       6 b.
D1.6       7 Cqb.
D3.5       8 c.
D1.8       9 Cqc.
D4.9      10 CCcqEqc.
D10.7     11 Ebc.

Answer 3

I assume you have a natural deduction system in mind:

Goal: Derive $B \leftrightarrow C$ from the set of premises $\{B \wedge C\}$, in symbols:

$B \wedge C \vdash B \leftrightarrow C.$

Recall that for deriving a biconditional $\phi \leftrightarrow \psi$ we first have to show that a conditional holds in both sides,

$\vdash \phi \rightarrow \psi$ and $\vdash \psi \rightarrow \phi.$

Hence, our new goal is derive both $B \rightarrow C$ and $C \rightarrow B$. Let's first obtain the former.

Since we want to obtain a conditional $B \rightarrow C$, we (i) assume the antecedent $B$ as a hypothesis, (ii) derive the consequent under this assumption, (iii) use the conditional introduction rule to conclude that $B$ implies $C$:

1. $B$, H
2. $B \wedge C$, P
3. $C$, 2 $\land$E (Rule called Conjunction Elimination)

4. $B \to C$, 1-3 $\to$I (Conditional Introduction)

Now we do the same to obtain $C \to B$

5. $C$, H
6. $B \wedge C$, P
7. $B$, 6 $\land$E (Conjunction Elimination)

8. $C \to B$, 5-7 $\to$I (Conditional Introduction)

and we use lines 4 and 8 to get a biconditional:

9. $C \leftrightarrow B$, 4, 8 $\leftrightarrow$I (Rule called Biconditional Introduction)

Now let's put all the steps together:

1. $B$, H
2. $B \wedge C$, P
3. $C$, 2 $\land$E

4. $B \to C$, 1-3 $\to$I

5. $C$, H
6. $B \wedge C$, P
7. $B$, 6 $\land$E

8. $C \to B$, 5-7 $\to$I
9. $C \leftrightarrow B$, 4, 8 $\leftrightarrow$I

Did you get the idea?