The proof that is shown. Given: ΔMNQ is isosceles with base , and and bisect each other at S.
Prove:

Square M N Q R is shown with point S in the middle. Lines are drawn from each point of the square to point S to form 4 triangles.

(1) Contradiction

(2) Tautology

(3) Neither Tautology nor Contradiction

(4) Tautology

Step-by-step explanation:

Given from the option that ,

(1) (P ↔ Q) ↔ ((P ∧ ¬Q) ∨ (¬P ∧ Q))

(2) (P → Q) V (Q →P)

(3) (¬R→((P V Q) → Q)) V R

(4) ((P→Q) ∧ (Q→R)) → (P→R)

From the first option (1), we have that;

(1)(P ↔ Q) ↔ ((P ∧ ¬Q) ∨ (¬P ∧ Q))

PQ¬P¬QP ↔ QP ∧ ¬Q¬P ∧Q(P∧¬Q) V (¬P∧Q)  result

TTFTFFFFF

TFFTFTFTF

FTTFFFTTF

FFTTTFFFF

From the result in the last column, this is a contradiction.

From option (2) we have that;  

(2)(P → Q) V (Q →P)

The resulting table becomes  

PQP →QQ → P  result

TTTTT

TFFTT

FTTFT

FFTTT

It is seen from the table that the result of the last column has all T which implies that it is a Tautology.

From option (3) we have that;

(3) (¬R→((P V Q) → Q)) V R

The resulting table becomes  

P  QR¬RPVQ(PVQ)→Q¬R→(PVQ→Q)Result

TTTFTTTT

TTFTTTTT

TFTFTFTT

TFFTTFFF

FTTFTTTT

FTFTTTTT

FFTFFTTT

FFFTFTTT

There is a little difference in the last column when compared to others,  

Here we can see that the last column contains all T but just one F, this implies that it is neither a Tautology nor Contradiction.

From option (4) we have that;

(4) ((P→Q) ∧ (Q→R)) → (P→R)

The resulting table becomes

PQRP →QQ →R(P→Q)∧(Q→R)P→R  result

TTT   T           T                 T             T             T

TTF   T           F                 F              F             T

TFT   F           T                 F              T             T

TFF   F           T                 F              F             T

FTT   T           T                 T               T     T

FTF   T           F                 F               T     T

FFT   T           T                 T                T     T

FFF   T           T                 T                 T     T

Here we can see that the last column contains all T which implies that it is a Tautology.

i hope this helps, cheers.