Download 1 APPENDIX Proof of Theorem 2.1 Proof: Since Vars(ϕi1) ∩ Vars

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1
A PPENDIX
Proof of Theorem 2.1
Proof: Since Vars(ϕi1 ) ∩ Vars(ϕi2 ) = ∅, ϕi1 and
ϕi2 are pairwise independent. Since Φi = ϕi1 ϕi2 ,
answer tuple r is made up of the elements which
come from both Vars(ϕi1 ) and Vars(ϕi2 ). If we remove
all variables in MCS r (Φi ) from the database, then Φi
becomes a counterfactual cause for answer tuple r.
If we further remove Vars(ϕi1 ) from the database,
Φi becomes unsatisfiable and r is removed from the
answer accordingly. Thus MCS r (ϕi1 ) = MCS r (Φi ).
With the similar method, we can derive MCS r (ϕi2 ) =
MCS r (Φi ). This proves the theorem.
Proof of Corollary 2.1
Proof: Since Φi = ϕi1 ϕi2 and Vars(ϕi1 ) ∩
Vars(ϕi2 ) = ∅, we can derive the following results:
1. MCS ϕi1 (xj ) is independent of ϕi2 and
MCS ϕi2 (yj ) is independent of ϕi1 . Thus, we can
independently compute MCS ϕi1 (xj ) and MCS ϕi2 (yj ).
2. Since MCS Φ (ϕi1 ) = MCS Φ (ϕi2 ) = MCS Φ (Φi ) by
Theorem 2.1, MCS Φi (ϕi1 ) = MCS Φi (ϕi2 ) = 0.
In order to compute MCS Φ (xj ), we should compute
MCS Φ (Φi ) first. Then we compute MCS Φi (ϕi1 ). Finally, we compute MCS ϕi1 (xj ). Thus, we can compute
MCS Φ (xj ) as follows:
MCS Φ (xj ) = MCS Φ (Φi )+MCS Φi (ϕi1 )+MCS ϕi1 (xj )
= MCS Φ (Φi ) + MCS ϕi1 (xj )
= MCS Φ (Φi ) + α
With the similar method, we can derive
MCS Φ (yi ) = MCS Φ (Φi ) + β. This proves the
corollary.
Proof of Proposition 3.1
Proof: From the lineage matrix, we observe that
Φ0 = LM [0][0]LM [0][1] · · · LM [0][n − 1] denotes a
set of clauses. Thus for any clause ci ∈ Φ, ci is
conjunction of one and only one variable from every column in its first row. By Theorem 2.1, we
can derive MCS r (Φ0 ) = MCS r (LM [0][0]) = · · · =
MCS r (LM [0][n − 1]). Thus we should not remove any
elements along the path from LM [0][0] to LM [0][j] in
order to compute MCS (LM [0 ][j ]). Hence SPM [0 ][j ] =
0. So the distance of the shortest path from P M [0][0]
to P M [0][j](j > 0) is 0. The proposition is proved.
Proof of Proposition 3.2
Proof: From the lineage matrix, we know that
∑l
Φ =
i=0 Xi fxi , where Xi denotes LM [i][0] and
fx0 ⊃∑· · · ⊃ fxl . We
can transform
∑i−1
∑lΦ as follows:
l
Φ =
X
f
=
X
f
+
k=0 k xk
k=i Xk fxk . Let
∑li=0 i xi
Φ =
X fxk . Since fx0 ⊃ · · · ⊃ fxl , MCS (Φ1 ) =
∑1i−1 k=i k ∑
i−1
|X
|
=
k
k=0
k=0 P M [k][0]. Hence SPM [k ][0 ] =
∑i−1
P
M
[k][0].
So
the distance of the shortest path
k=0
∑i−1
from P M [0][0] to P M [i][0](i > 0) is k=0 P M [k][0].
This proves the proposition.
Proof of Proposition 3.3
Proof: From the path matrix, we observe that if
P M [i][j − 1] is not a node then P M [0][0] can not
reach P M [i][j] by way of P M [i][j − 1]. So the shortest
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path from P M [0][0] to P M [i][j] does not include cell
P M [i][j − 1]. Thus, SPM [i][j] has nothing to do with
P M [i][j − 1]. With the similar reason, SPM [i][j] has
no concern with P M [i − 1][j] if P M [i − 1][j] is not a
node.
When computing SPM [i][j], we have computed
SPM [i − 1][j] and SPM [i][j − 1] because we compute
SPM [m][n] from the first row to the last row and in
every row from the first column to the last column.
Since path matrix is directed, P M [0][0] can reach cell
P M [i][j] by way of cells P M [i − 1][j] and P M [i][j − 1]
if P M [i − 1][j] and P M [i][j − 1] are both nodes. Thus
SPM [i][j] is related to SPM [i − 1][j] and SPM [i][j − 1]
if P M [i − 1][j] and P M [i][j − 1] are both nodes.
Let ϕj = LM [0][j] for 0 ≤ j ≤ n − 1 and Φ0 = ϕ0 ∧
ϕ1 · · · ∧ ϕn−1 . From the second row of lineage matrix
LM [m][n], if LM [i][j] is not equal to 0, then ϕj =
LM [i][j]. After computing all ϕj s, Φi = ϕ0 ∧ ϕ1 · · · ∧
ϕn−1 . By Theorem 2.1, we can derive MCS r (Φi ) =
MCS r (ϕ0 ) = MCS r (ϕ1 ) = · · · = MCS r (ϕn−1 ). Thus, if
P M [i][j − 1] is a node, SPM [i][j] = SPM [i][j − 1] from
direction P M [i][j − 1] → P M [i][j].
However, if P M [i−1][j] is a node, Vars(LM [i−1][j])
and Vars(LM [i][j]) first appear in different clauses.
Thus SPM [i][j] ̸= SPM [i − 1][j]. Since P M [0][0] must
pass through one more node P M [i − 1][j] to reach
P M [i][j] from direction P M [i − 1][j] → P M [i][j],
SPM [i][j] = SPM [i − 1][j] + P M [i − 1][j].
Let x = SPM [i][j − 1] and y = SPM [i − 1][j] +
P M [i−1][j]. Combining two directions P M [i−1][j] →
P M [i][j] and P M [i][j − 1] → P M [i][j], we know that
SPM [i][j] = min(x, y) for i > 0. If either x or y is
equal to 0, its corresponding cell is not a node. Thus
we delete it from Function min(). This proves the
proposition.
Proof of Lemma 3.1
Proof: Let ϕj = LM [0][j] for 0 ≤ j ≤ n − 1 and
Φ0 = ϕ0 ∧ϕ1 · · ·∧ϕn−1 . From the second row of lineage
matrix LM [m][n], if LM [i][j] is not equal to 0, then
ϕj = LM [i][j]. After computing all ϕj s, Φi = ϕ0 ∧
ϕ1 · · · ∧ ϕn−1 . Then,
1. If a directed edge from LM [i][j] passes through
k rows, then LM [i][j] influences k sets of clauses Φi s.
If all variables in LM [i][j] are set to false, then the
k sets of clauses become false. The mapping path
from LM [0][0] to LM [i][j] must include directed edges
which add up to passing through the first i rows.
Thus, if all variables in the cells in the first i rows
along the mapping path from LM [0][0] to LM [i][j]
are set to false, all clauses including any variable in
colj (Φ) − Vars(LM [i ][j ]) − col[j] become false.
2. If all variables in any col[k](j ≤ k < n) are set to
false, all variables in cells LM [l][k](i < l < n) become
false. Since a clause is conjunction of variables from
all columns, all clauses including any variable in col[j]
become false.
This proves the lemma.
Proof of Theorem 3.1
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Proof: By Lemma 3.1 Item 1, all clauses including
any variable in colj (Φ)−Vars(LM [i ][j ])−col[j] become
false if all variables in the cells in the first i rows along
the mapping path from LM [0][0] to LM [i][j] are set to
false. Let col[k] have the minimal number of elements
among the sets col[j], . . . , col[n−1]. By Lemma 3.1 Item
2, all clauses including any variables in col[j] become
false if all variables in col[k] are set to false.
After the above two sets of variables are set to
false, only clauses including variables in LM [i][j]
remain. Thus those clauses become actual cause
with respect to Φ. Let minValue = |col [k]| =
min(|col[j]|, . . . , |col[n − 1]|). So MCS Φ (LM [i][j]) =
SPM [i ][j ] + minValue. Since P M [i][j] = |LM [i ][j ]|,
resp(xk ) = SPM [i][j ]+P M1[i][j]+minValue for any xk ∈
LM [i ][j ]. This proves the theorem.
Proof of Lemma 3.2
Proof: Let Φi = Xi fxi = Xi fyi fzi . Since Xi is
conjunction with fyi and fzi to get Φi , we can not
compute responsibility of any variable in Xi by performing responsibility analysis for Φi1 = Xi alone. For
formula Φ′ = fyi fzi , we can independently compute
responsibility of any variable in fyi and fzi by Corollary 2.1 because Vars(fyi ) ∩ Vars(fzi ) = ∅. Since Xi is
conjunction with fyi and fzi to get Φi , we can compute
responsibility of any variable in fyi by performing
responsibility analysis for Φi2 = Xi fyi alone. With the
similar reason, we can compute responsibility of any
variable in fzi by performing responsibility analysis
for Φi3 = Xi fzi alone.
Let Φj = Xj fxj = Xj fyj fzj (j ̸= i). Computing
responsibility of any variable in fyj is independent of fzj by performing responsibility analysis for
Φj2 = Xj fyj alone. Since Vars(fyj ) ∩ Vars(fzi ) = ∅,
computing responsibility of any variable in fyj is also
independent of fzi . So, computing responsibility of
∑l
∑l
any variable in i=1 fyi is independent of j=1 fzj
by performing responsibility analysis for Φ1 alone.
Thus, resp Φ (yi ) = resp Φ1 (yi ) for any yi ∈ Φ1 . With
the similar reason, computing responsibility of any
∑l
∑l
variable in
j=1 fzj is independent of
i=1 fyi by
performing responsibility analysis for Φ2 alone. Thus,
resp Φ (zi ) = resp Φ1 (zi ) for any zi ∈ Φ2 .
Finally, for any xi ∈ Φ, we get MCS Φ1 (xi )
and MCS Φ2 (xi ) by performing responsibility
analysis for Φ1 and Φ2 , respectively. Thus,
MCS Φ (xi ) = min(MCS Φ1 (xi ), MCS Φ2 (xi )) and
resp Φ (xi ) = max(resp Φ1 (xi ), resp Φ2 (xi )). This proves
the lemma.
Proof of Lemma 3.3
Proof: For any two path lineages Φi and Φj (1 ≤
i, j ≤ k), Vars(Φi ) ∩ Vars(Φj ) = {X1 , . . . , Xl }. By
Lemma 3.2, we can compute responsibilities of all
variables in Φi ∪ Φj by independently performing
responsibility analysis for Φi , Φj and resp Φi ∪Φj (xl ) =
max(resp Φi (xl ), resp Φj (xl )) for any xl ∈ Φ. Since Φ is decomposed into Φ1 , . . . , Φk , resp Φ (xi ) =
max(resp Φ1 (xi ), . . . , resp Φk (xi )) for any xi ∈ Φ. This
2
proves the lemma.
Proof of Theorem 3.3
Proof: Assume that the first split node is Y .
Then Φ is decomposed into a set of path lineages
Φ1 , . . . , Φk−1 and a smaller composite lineage Φk . By
Lemma 3.3, we can use Algorithm 1 to perform responsibility analysis for Φ1 , . . . , Φk−1 and resp Φ (xi ) =
max(resp Φ1 (xi ), . . . , resp Φk−1 (xi ), resp Φk (xi )) for any
xi ∈ Φ.
Then we split the inequality graph of Φk recursively
until it is split into a set of inequality paths. In each
step, we employ Algorithm 1 to perform responsibility analysis for the corresponding path lineages.
Thus we can decompose Φ into path lineages for
responsibility analysis. This proves the theorem.