DHARM
PROPOSITIONAL LOGIC 133
Now construct the tableaux for this formula
∼→∧→(((A B) A) B)
A
~A B
Moving according to this path
we get a contradiction (A & A)∼
moving according to this path we get
we get a contradiction (B & B)∼
(A→∧B) A
~B
(A→B)
××
So, both paths in the tableaux are closed, therefore tableaux is closed. It follows the
formula (~ X) labeled at the root is a contradiction. Therefore, X is a tautology and so argument
is a valid argument.
Example 5.27. Show that
1.M → J
2.J → ~ H
- ~ H → ~ T
- ~ T → M
5.M → ~ H /∴ ~ T
Sol. Assume premises 1, 2, 3, 4, 5 are denoted by X 1 , X 2 , X 3 , X 4 , X 5 respectively. So the argu-
ment has the formula (say X), where
X : (((((X 1 ∧ X 2 ) ∧ X 3 ) ∧ X 4 ) ∧ X 5 ) → ~ T);
So, ~ X : ~ (((((X 1 ∧ X 2 ) ∧ X 3 ) ∧ X 4 ) ∧ X 5 ) → ~ T);
Fig. 5.30 shows the tableaux for ~ X. In which we find one open and complete (since all
formulas in this path are expended) path so tableaux is not closed. It implies that formula at
root ~ X is not a contradiction or X is a contradiction. Therefore, X is not a tautology and so
argument is invalid.
