Lossless vs Lossy Decomposition for Given Functional Dependencies (D1 and D2)

D1 fragments: $R_1(P,Q,S,T)$, $R_2(P,T,X)$, $R_3(Q,Y)$, $R_4(Y,Z,W)$. Common attributes among them are: $R_1\cap R_2=\{P,T\}$, $R_1\cap R_3=\{Q\}$, $R_3\cap R_4=\{Y\}$, and $R_2\cap R_4=\emptyset$.

Create one row per fragment. Use a distinguished symbol $a_P$ wherever an attribute appears in the first row fragment (the “seed”), otherwise use unique $b$-symbols for each attribute occurrence not in the seed row. (This is the standard chase-tableau construction for lossless join tests 2.)

Take the seed row as $R_1(P,Q,S,T)$ (so $P,Q,S,T$ are marked with $a$-symbols in that row; $X,Y,Z,W$ use fresh $b$-symbols in that row).

Use the chase rule: for each FD $U\to V$, if in some column(s) of $U$ two rows have the same “kind” of symbol (an $a$ matching an $a$, or a forced equality already established), then the corresponding $V$-columns must be equated as well.

Since $Q$ is present in $R_1$ and $R_3$, the $Q$-columns in the seed row and the $R_3$ row share the same determinant (seed has $a_Q$, and $R_3$ has $a_Q$ for $Q$). Applying $Q\to Y$ forces equality of the $Y$-columns between the seed row and the $R_3$ row (so the $Y$ symbols align).

Once $Y$ is aligned between the seed and the $R_3$ row, FD $Y\to ZW$ forces the $Z$ and $W$ columns to align in the $R_4$ row with the same symbol types as required by the tableau.

Because $P$ is in both $R_1$ and $R_2$, the tableau has aligned $P$ columns between the seed and $R_2$. Applying $P\to YX$ forces both $Y$ and $X$ columns in $R_2$ to align with the already-consistent $Y$ (from the $Q\to Y$ chain) and to produce consistent $X$.

After these propagation steps, the chase tableau yields at least one row where all attributes of the seed component match consistently (no contradictions/spurious symbol merging). Therefore the join of the fragments reproduces $R$ exactly, so D1 is lossless.

Overlaps: $R_1'\cap R_3' = \{Q\}$, $R_3'\cap R_4' = \{Y\}$, and there is no overlap between $R_2'(T,X)$ and $R_1'(P,Q,S)$ (they share none).

Use a standard chase tableau with one row per fragment per the lossless join test method 2. Choose seed row as $R_1'(P,Q,S)$.

Because $Q$ appears in both $R_1'$ and $R_3'$, chase applies $Q\to Y$ so $Y$ becomes consistent across those rows; then $Y\to ZW$ propagates consistency into $R_4'$.

To constrain $X$ in $R_2'(T,X)$, the tableau would need some FD whose LHS is present in the already-consistent components and whose RHS includes $X$. Candidate FDs with $X$ on RHS are: $PQ\to X$ and $P\to YX$.

In D2, $(T,X)$ shares no attributes with the seed component $(P,Q,S)$ or with $(Q,Y)$ / $(Y,Z,W)$. Therefore in the chase tableau, there is no way to align the determinant of $PQ\to X$ (requires both $P$ and $Q$ in the same intersecting structure) with rows that also contain $X$. Similarly, $P\to YX$ requires alignment of $P$, but $P$ never appears in the $R_2'(T,X)$ row, so the RHS symbol $X$ in that row never gets forced to match the $X$ symbols derived elsewhere.

Because the determinant information cannot reach the $X$-containing fragment through overlaps, the chase tableau does not collapse to a fully consistent consistent row. Hence the decomposition introduces spurious combinations upon joining, so D2 is lossy.