Verify if the following compound proposition is valid or not using the truth table.

Let LHS = (A∧B)∧C ⇒ (A∧B∧C), and RHS = (A∧B) ⇔ (A∧B∧C).

A	B	C	(A∧B)∧C	A∧B∧C	LHS	A∧B	RHS
0	0	0	0	0	1	0	1
0	0	1	0	0	1	0	1
0	1	0	0	0	1	0	1
0	1	1	0	0	1	0	1
1	0	0	0	0	1	0	1
1	0	1	0	0	1	0	1
1	1	0	0	0	1	1	0
1	1	1	1	1	1	1	1

LHS is of the form P⇒P, so it is always TRUE (a tautology, column is all 1's). RHS, however, is FALSE when A=1,B=1,C=0 (A∧B=1 but A∧B∧C=0), so RHS is not always true. Since LHS ≠ RHS for A=1,B=1,C=0 (row 7), the given compound proposition is NOT a tautology - it is a contingency. Hence the compound proposition is NOT VALID (not always true).