SOLUTION: I would like to know how can I solve below. F / (G->H) v (~G->J)

Question 1167106: I would like to know how can I solve below.
F / (G->H) v (~G->J)
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
This problem involves evaluating the truth value of the compound expression $F \lor (G \to H) \lor (\sim G \to J)$.
The key to solving this is understanding the truth tables for **disjunction** ($\lor$), **implication** ($\to$), and **negation** ($\sim$).
## Solving the Expression
The expression you provided is:
$$F \lor (G \to H) \lor (\sim G \to J)$$
### 1. The Disjunction Rule ($\lor$)
The entire expression is a large **disjunction** (OR statement) connecting three main clauses:
$$\text{CLAUSE}_1 \lor \text{CLAUSE}_2 \lor \text{CLAUSE}_3$$
The fundamental rule for disjunction is:
$$\text{A disjunction is **TRUE** if at least one of its components is **TRUE**.} \quad (T \lor X = T)$$
Therefore, if **any** of the three main clauses is true, the entire expression is **TRUE**, regardless of the truth values of the other clauses.
### 2. The Implication Rule ($\to$)
The two complex clauses use the **implication** (IF...THEN) operator.
$$\text{A} \to \text{B} \text{ is **FALSE** only when the antecedent (A) is **TRUE** and the consequent (B) is **FALSE**.} \quad (T \to F = F)$$
For all other combinations, the implication is **TRUE**.
### 3. Analysis by Case
To see if the entire expression can ever be false, we must check if it's possible for **all three clauses to be false simultaneously.**
$$\underbrace{F}_{\text{Clause 1}} \lor \underbrace{(G \to H)}_{\text{Clause 2}} \lor \underbrace{(\sim G \to J)}_{\text{Clause 3}} = F$$
This requires:
1. **Clause 1 is False:** $F$ must be **False**.
$$F = F$$
2. **Clause 2 is False:** $G \to H$ must be **False**.
This only happens when the antecedent is True and the consequent is False:
$$G = T \quad \text{and} \quad H = F$$
3. **Clause 3 is False:** $\sim G \to J$ must be **False**.
This only happens when the antecedent ($\sim G$) is True and the consequent ($J$) is False:
$$\sim G = T \quad \text{and} \quad J = F$$
### 4. Checking for Consistency
Let's combine the requirements from all three clauses:
* From Clause 1: $F = F$
* From Clause 2: $G = T$
* From Clause 2: $H = F$
* From Clause 3: $J = F$
* From Clause 3: $\sim G = T$
We have a **contradiction** because one condition requires $G = T$ and another requires $\sim G = T$ (which means $G = F$). It is logically impossible for both $G$ to be True and $G$ to be False simultaneously.
Since it is **impossible** for all three clauses to be false, the entire expression must always be **TRUE**.
## Conclusion
The expression $F \lor (G \to H) \lor (\sim G \to J)$ is a **tautology**; it is **always True** for any combination of truth values for $F$, $G$, $H$, and $J$.
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