Theorem D.2 (ECHOIs Arbitrary) The computer programECHO, which mod-els the theory of explanatory coherenceT EC of Thagard (1989), is arbitrary.
Proof.
The definition of the measureH(S, t)of the global coherence of a systemS ofn propositions at timetruns as follows:
H(S, t) = X
0≤i≤n
X
0≤j≤n
wij ·ai(t)·aj(t).
wij is the weight of the excitatory or inhibitory link from unitito unitj,ai(t)is the activation of unitiat timet, andnis the number of propositions in the system Swhich are represented by the units1, . . . , n.
An excitatory link between two unitsiandjrepresents a coherence relation between the two propositions the units i and j stand for, whereas an inhibitory link represents an incoherence relation. The activation ai(t) of unit i at time t expresses the degree of acceptance of the proposition represented by unitiat time t.
An input in form of (values for the) activationsai(0)of the unitsiof some system of propositions S at time 0 is used to set up a network which includes – besides the units 1, . . . , n – a special unit 0 with activation a0(t) = 1, for every timet. Then the network is run in cycles that synchronously update all the units so that the activation streams from the special unit0over units representing data (evidences) to units representing hypotheses which are explanatorily linked to these data.
The activation ai(·), ai(·) : N → [−1,1], of any unit i is a continuous function of all unitsj linked to it. The contribution of each such unit j depends on the weight wij of the link from i to j. These weights wij – expressing the strength of the (in)coherence relation between the propositionsPk andQ, which are represented by the unitsiandj, respectively – have to obey the equation
weight(Pk, Q) = default weight
(number of cohypotheses of Pk)simplicity impact, whereQis explained byP1, . . . , Pk, . . . , Pm,1≤k ≤m.
Despite this, thewij can be chosen in an arbitrary way, for both the default weight and the simplicity impact can be freely chosen. The number of cohy-potheses of propositionPk is the numberm−1of propositions that occur in the
explanation of Q by P1, . . . , Pk, . . . , Pm apart from Pk. The activation ai(·) of unitiis updated by the following equation:
ai(t+ 1) =
ai(t)·(1−θ) +neti(t+ 1)·(max−ai(t)), ifneti(t+ 1)>0,
ai(t)·(1−θ) +neti(t+ 1)·(ai(t)−min), ifneti(t+ 1)≤0.
θ is a decay parameter decrementing each unitiat every cycle, min = −1is the minimum activation,max = 1is the maximum activation, andneti(t+ 1) is the net input to unitiat timet+ 1, which is given as
neti(t+ 1) = X
0≤j≤n
wij ·aj(t),
wherenis again the number of propositions (respectively units without the special unit0) in the system of propositionsS. By repeating updating cycles some units get activated, whereas others get deactivated.
In order to prove the above claim it suffices to give an example of two sets of propositions S1 and S2, and two measures H(·,·) and H0(·,·) of the global coherence of a system of propositions for which there is a time t0 such that it holds for every timet≥t0:
H(S1, t)> H(S2, t) and H0(S1, t)< H0(S2, t).
LetS1 ={E1, P2, P3}, where evidenceE1is supposed to be explained by each of the two hypotheses P2 andP3. E1 is represented by1, P2 by2, andP3 by3; the special unit with activation1is represented by0. Let
w10=w01= 1, w12=w21 = 1 =w13 =w31, θ = 0.
Thus the strength of the explanatory relation between P2 and E1 is assumed to be equal to the strength of the explanatory relation between P3 andE1; and both are supposed to be equal to the degree of acceptance whichE1 has qua being an evidence.
a1(1) =a1(0)·(1−θ) +net1(1)·(max−a1(0)) = 0·(1−0) + 1·(1−0) = 1, for
net1(1) = w10·a0(0) +w12·a2(0) +w13·a3(0) = 1·1 + 1·0 + 1·0 = 1>0.
a2(1) =a2(0)·(1−θ)+net2(1)·(a2(0)−min) = 0·(1−0)+0·(0−(−1)) = 0, for
net2(1) =w21·a1(0) = 1·0 = 0≤0.
a3(1) =a3(0)·(1−θ)+net3(1)·(a3(0)−min) = 0·(1−0)+0·(0−(−1)) = 0, for
net3(1) =w31·a1(0) = 1·0 = 0≤0.
a1(2) =a1(1)·(1−θ) +net1(2)·(max−a1(1)) = 1·(1−0) + 1·(1−1) = 1, for
net1(2) =w10·a0(1) +w12·a2(1) +w13·a3(1) = 1·1 + 1·0 + 1·0 = 1>0.
a2(2) =a2(1)·(1−θ) +net2(2)·(max−a2(1)) = 0·(1−0) + 1·(1−0) = 1, for
net2(2) =w21·a1(1) = 1·1 = 1 >0.
a3(2) =a3(1)·(1−θ) +net3(2)·(max−a3(1)) = 0·(1−0) + 1·(1−0) = 1, for
net3(2) =w31·a1(1) = 1·1 = 1 >0.
So
H1(S1,2) = X
0≤i≤3
X
0≤j≤3
wij ·ai(2)·aj(2)
= w01·a0(2)·a1(2) +w10·a1(2)·a0(2) + +w12·a1(2)·a2(2) +w13·a1(2)·a3(2) + +w21·a2(2)·a1(2) +w31·a3(2)·a1(2)
= 1·1·1 + 1·1·1 + 1·1·1 + +1·1·1 + 1·1·1 + 1·1·1
= 6.
It will be shown (by induction on timet) thatai(t) = 1, for everyiandt,1≤i≤ 3, t ≥2. Lett≥2, and suppose the induction hypothesis holds.
a1(t+ 1) = a1(t)·(1−θ) +net1(t+ 1)·(max−a1(t)) net1(t+ 1) = 3>0
= 1·(1−θ) +net1(t+ 1)·(max−1) by induction hypothesis
= 1·(1−0) + 3·0 θ = 0, net1(t+ 1) = 3
= 1,
for
net1(t+ 1) = w10·a0(t) +w12·a2(t) +w13·a3(t)
= w10·1 +w12·1 +w13·1 by induction hypothesis
= 1·1 + 1·1 + 1·1 = 3 >0.
a2(t+ 1) = a2(t)·(1−θ) +net2(t+ 1)·(max−a2(t)) net2(t+ 1) = 1>0
= 1·(1−θ) +net2(t+ 1)·(max−1) by induction hypothesis
= 1·(1−0) + 0·(1−1) θ = 0, net2(t+ 1) = 1
= 1, for
net2(t+ 1) = w21·a1(t)
= w21·1 by induction hypothesis
= 1·1 = 1>0.
a3(t+ 1) = a3(t)·(1−θ) +net3(t+ 1)·(max−a3(t)) net3(t+ 1) = 1>0
= 1·(1−θ) +net3(t+ 1)·(max−1) by induction hypothesis
= 1·(1−0) + 0·(1−1) θ = 0, net3(t+ 1) = 1
= 1, for
net3(t+ 1) = w31·a1(t)
= w31·1 by induction hypothesis
= 1·1 = 1>0.
Thus for everyt≥2:
H1(S1, t) = X
0≤i≤3
X
0≤j≤3
wij ·ai(t)·aj(t)
= w01·a0(t)·a1(t) +w10·a1(t)·a0(t) + +w12·a1(t)·a2(t) +w13·a1(t)·a3(t) + +w21·a2(t)·a1(t) +w31·a3(t)·a1(t)
= 1·1·1 + 1·1·1 + 1·1·1 + +1·1·1 + 1·1·1 + 1·1·1
= 6.
LetS2 ={E1, E2},E1 andE2being evidences. E1 is represented by1, E2 by2, and the special unit with activation1is represented by0. Let
w10 =w01=w20 =w02= 1, θ= 0.
Thus the degree of acceptance whichE1 has qua being an evidence is supposed to be equal to the degree of acceptance whichE2has qua being an evidence.
a1(1) =a1(0)·(1−θ) +net1(1)·(max−a1(0)) = 0·(1−0) + 1·(1−0) = 1, for
net1(1) =w10·a0(0) = 1·1 = 1 >0.
a2(1) =a2(0)·(1−θ) +net2(1)·(max−a2(0)) = 0·(1−0) + 1·(1−0) = 1, for
net2(1) =w20·a0(0) = 1·1 = 1 >0.
a1(2) =a1(1)·(1−θ) +net1(2)·(max−a1(1)) = 1·(1−0) + 1·(1−1) = 1, for
net1(2) =w10·a0(1) = 1·1 = 1 >0.
a2(2) =a2(1)·(1−θ) +net2(2)·(max−a2(1)) = 1·(1−0) + 1·(1−1) = 1, for
net2(2) =w20·a0(1) = 1·1 = 1 >0.
So
H1(S2,2) = X
0≤i≤2
X
0≤j≤2
wij ·ai(2)·aj(2)
= w01·a0(2)·a1(2) +w02·a0(2)·a2(2) + +w10·a1(2)·a0(2) +w20·a2(2)·a0(2)
= 1·1·1 + 1·1·1 + 1·1·1 + 1·1·1
= 4.
It will be shown (by induction on timet) thatai(t) = 1, for everyiandt,1≤i≤ 2, t≥2. Lett≥2, and suppose the induction hypothesis holds.
a1(t+ 1) = a1(t)·(1−θ) +net1(t+ 1)·(max−a1(t)) net1(t+ 1) = 1>0
= 1·(1−θ) +net1(1)·(max−1) by induction hypothesis
= 1·(1−0) + 1·(1−1) θ = 0, net1(t+ 1) = 1
= 1, for
net1(t+ 1) =w10·a0(t) = 1·1 = 1>0.
a2(t+ 1) = a2(t)·(1−θ) +net2(t+ 1)·(max−a2(t)) net2(t+ 1) = 1>0
= 1·(1−θ) +net2(t+ 1)·(max−1) by induction hypothesis
= 1·(1−0) + 1·(1−1) θ = 0, net2(t+ 1) = 1
= 1, for
net2(t+ 1) =w20·a0(t) = 1·1 = 1>0.
So it holds for everyt≥2:
H1(S2, t) = X
0≤i≤2
X
0≤j≤2
wij ·ai(t)·aj(t)
= w01·a0(t)·a1(t) +w02·a0(t)·a2(t) + +w10·a1(t)·a0(t) +w20·a2(t)·a0(t)
= 1·1·1 + 1·1·1 + 1·1·1 + 1·1·1
= 4.
It follows for everyt≥2:
H1(S1, t) = 6>4 =H1(S2, t).
Consider againS1 ={E1, P2, P3}, where evidenceE1is supposed to be explained by each ofP2andP3. This time let
w010=w010 = 1, w012=w210 = 1/10 =w130 =w310 , θ0 = 0.
Thus the strength of the explanatory relation betweenP2andE1 is again assumed to be equal to the strength of the explanatory between P3 and E1; but this time they are both supposed to be smaller than the degree of acceptance whichE1 has qua being an evidence.
a1(1) =a1(0)·(1−θ) +net01(1)·(max−a1(0)) = 0·(1−0) + 1·(1−0) = 1, for
net01(1) = w010·a0(0) +w120·a2(0) +w013·a3(0)
= 1·1 + (1/10)·0 + (1/10)·0 = 1>0.
a2(1) =a2(0)·(1−θ)+net02(1)·(a2(0)−min) = 0·(1−0)+0·(0−(−1)) = 0, for
net02(1) =w021·a1(0) = (1/10)·0 = 0≤0.
a3(1) =a3(0)·(1−θ)+net03(1)·(a3(0)−min) = 0·(1−0)+0·(0−(−1)) = 0, for
net03(1) =w031·a1(0) = (1/10)·0 = 0≤0.
a1(2) =a1(1)·(1−θ) +net01(2)·(max−a1(1)) = 1·(1−0) + 1·(1−1) = 1, for
net01(2) = w010·a0(1) +w120·a2(1) +w013·a3(1)
= 1·1 + (1/10)·0 + (1/10)·0 = 1>0.
a2(2) = a2(1)·(1−θ) +net02(2)·(max−a2(1))
= 0·(1−0) + (1/10)·(1−0) = 1/10, for
net02(2) =w021·a1(1) = (1/10)·1 = 1/10>0.
a3(2) = a3(1)·(1−θ) +net03(2)·(max−a3(1))
= 0·(1−0) + (1/10)·(1−0) = 1/10, for
net03(2) =w031·a1(1) = (1/10)·1 = 1/10>0.
So degree of acceptance whichE1has qua being an evidence is supposed to be equal to the degree of acceptance whichE2 has qua being an evidence. Again, let
w010=w010 =w020=w002= 1, θ0 = 0.
As
w010=w10, w001=w01, w020 =w02, w020=w20, θ0 =θ,
it follows that
H0(S2,2) =H(S2,2) = 4,
and – by the same reasoning (induction) as above – that it holds for everyt≥2:
H0(S2, t) =H(S2, t) = 4.
Thus
H0(S1, t) = 2.04<4 = H0(S2, t), for everyt ≥2.
Put together these results yield that there is at least onet0 (anyt0 ≥2does the job) such that it holds for everyt≥t0:
H(S1, t)> H(S2, t) and H0(S1, t)< H0(S2, t).
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