order-preserving. Since H has only finitely many
minimal elements, one of them must correspond to the
maximum of M, thus S is inversely well ordered.
Due to the previous lemma, we can introduce the
following residua on S:
a → b = max{z ∈ S | z⋆ a ≤ b},
a b = max{z ∈ S | a⋆ z ≤ b}.
Therefore we can prove the following result:
Proposition 3.1. The enriched monoid
S = (S,⋆,→, ,≤, 0,1)
is a psMTL
r
-chain and there exists an evaluation e
S
of T such that e
S
(ϕ) < 1.
Proof. We know that S is an integral totally ordered
monoid and that ≤ is compatible with ⋆. The fact that
a⋆ b ≤ c iff a ≤ b → c iff b ≤ a c follows immedi-
ately from the definitions of → and . Hence S is an
psMTL
r
-chain.
Let us define an evaluation e
S
(p) = e
L
(p), for
every propositional variable p appearing in any χ ∈
T ∪ {ϕ} and arbitrarily, otherwise. Let χ ∈ T ∪ {ϕ}.
We show by induction on the complexity of χ that
e
S
(ψ) = e
L
(ψ), for each subformula ψ of χ. Al-
most all cases are trivial, therefore we treat only the
case when ψ = ψ
1
ψ
2
. By definition we have
e
S
(ψ) = e
L
(ψ
1
) e
L
(ψ
2
) = max{z ∈ S | e
L
(ψ
1
)⋆z ≤
e
L
(ψ
2
)}. Since ψ is a subformula of χ, we have
e
L
(ψ) ∈ S. Consequently, e
L
(ψ
1
)⋆ e
L
(ψ) = e
L
(ψ
1
)∗
L
e
L
(ψ) ≤ e
L
(ψ
2
). Thus e
L
(ψ) ≤ e
S
(ψ). Now sup-
pose there is an element z
′
∈ S such that z
′
> e
L
(ψ)
and e
L
(ψ
1
) ⋆ z
′
≤ e
L
(ψ
2
). Since z
′
∈ L, we get z
′
≤
e
L
(ψ
1
) e
L
(ψ
2
) = e
L
(ψ) (contradiction). Hence
e
S
(ψ) = e
L
(ψ).
Since e
S
(τ) = e
L
(τ) = 1, for any τ ∈ T, e
S
is an
S-model of T. Moreover, e
S
(ϕ) = e
L
(ϕ) < 1.
Therefore we have the psMTL
r
-chain S which is
countable and inversely well ordered, and the evalu-
ation e
S
on S such that e
S
(ϕ) < 1. The next step is
to build a new psMTL
r
-chain S
′
order-isomorphic to
[0,1] in which S can be embedded.
In order to define such a psMTL
r
-chain, we will
use a similar construction as in the original proof of
the standard completeness theorem from (Jenei and
Montagna, 2003). We define the new universe by:
S
′
= {(a,x) | a ∈ S \{0}, x ∈ (0,1]} ∪{(0,1)}.
The order ≤
′
on S
′
is the lexicographic order, i.e.
(a,x) ≤
′
(b,y) iff a < b or (a = b and x ≤ y).
Let I be the set of all idempotents of S, i.e x⋆ x = x.
We define the following monoidal operation on S
′
:
(a,x)⋆
′
(b,y) =
(a⋆ b,1), a⋆ b < a∧ b
(a,xy), a = b, a ∈ I
min{(a,x),(b, y)}, otherwise
where xy stands for the usual product of reals. Note
that in any case, the first coordonate always equals
a⋆ b.
The proof of the following result follows closely
the proof of Lemma 6 from (Horˇcik, 2007), but dif-
ferences do appear since we are dealing with a non-
commutativity operation ⋆
′
.
Lemma 3.2. The structure
S
′
= (S
′
,⋆
′
,≤
′
,(0,1), (1,1))
is a totally ordered integral monoid, where (1,1) is
the neutral element and the top element as well, (0,1)
is the bottom element and ⋆
′
is monotone with respect
to ≤
′
on both arguments.
Proof. It is obvious that (1,1) is the neutral element
and the top element and that (0, 1) is the bottom ele-
ment. Let us prove that ⋆
′
is associative, i.e.
(a,x) ⋆
′
((b,y) ⋆
′
(c,z)) = ((a,x) ⋆
′
(b,y)) ⋆
′
(c,z).
We denote the left-hand side (right-hand side, respec-
tively) of the above equation by L (R, respectively).
Let P(a,b) denote the following property of a,b ∈ S
′
:
P(a,b) : a = b and a ∈ I.
Clearly, if P(a,b) holds, then P(b,a) holds as well.
We must analyse several cases:
1. Suppose that none of P(a,b), P(b,c), P(a,b ⋆ c),
P(a⋆ b,c) is valid. We distinguish two cases:
(i) a⋆b⋆ c = a∧ b∧ c. Then both L and R are equal
to min{(a,x),(b, y),(c,z)} and therefore associa-
tivity of ⋆
′
holds.
(ii) a ⋆ b ⋆ c 6= a∧ b ∧ c. We claim that in this case
both L and R are equal to (a⋆ b⋆ c,1). Let us con-
sider the case of L, since for R we have a similar
proof. If b⋆ c = b∧ c, then a⋆ (b⋆ c) 6= a∧(b⋆ c),
therefore L = (a ⋆ b⋆ c,1). If b ⋆ c 6= b ∧ c, then
(b,y) ⋆
′
(c,z) = (b ⋆ c,1) and a ⋆ b⋆ c 6= a. Thus
either a⋆ b ⋆ c 6= b ⋆ c and then L = (a ⋆ b ⋆ c,1),
or a⋆ b⋆ c = b⋆ c and then L = (a,x)∧ (b⋆c, 1) =
(b⋆ c,1) = (a⋆ b⋆ c,1).
2. Suppose that P(b,c) holds. Then b = c, b is idem-
potent and L = (a,x) ⋆
′
(b,yz). We obtain that
L =
(b,xyz), if a = b
(a⋆ b,1), if a ⋆ b < a ∧ b
(a,x), if a⋆ b = a∧ b,a < b
(b,yz), if a⋆ b = a∧ b,a > b
.
For R we have the following cases:
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