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JOURNAL OF COMPUTERS, VOL. 8, NO. 6, JUNE 2013 1439
L
L
Let v = v = a , φ
M, v M,
w
M ,
= b
w w
, thus we must
prove inf
w( a⇒bw) ≤( a⇒ inf
wbw)
. On the one hand,
inf w
b w
≤ b w
, thus a⇒bw ≥( a⇒ inf
wbw)
for each w,
thus inf
w( a⇒bw) ≥( a⇒ inf
wbw)
. On the other hand if
z ≤( a⇒ b w
) for each w, then z∗a ≤ bw
for each w,
z∗a ≤ inf w
b w
, z ≤( a⇒ inf
wbw)
. Thus ( a⇒ inf
wbw)
is
the infimum of all ( a⇒
b w
). So (U19) holds.
For (U20), we need to verify
inf
w( aw ⇒ bw) = (sup
w
aw
⇒ b)
. Indeed, sup w
≥ a w
, thus
(sup
w
aw ⇒b) ≤( aw
⇒ b)
, hence
(sup
w
aw ⇒b) ≤inf w( aw
⇒ b)
, If z ≤ aw
⇒ b for all w ,
then aw
≤( z ⇒ b)
for all w, then sup
w
aw
≤( z ⇒ b)
,
z ≤(sup waw
⇒ b)
, so sup w
a w
⇒ b is the infimum. So
(U20) holds.
Finally we verify (U21), we need to verify
inf
w( a∨ bw) = a∨ infwbw
. Indeed, a ≤ a∨ bw
, thus
a ≤inf w( a∨ bw)
; similarly, infwbw ≤inf w( a∨ bw)
, thus
a∨infwbw ≤inf w( a∨ b)
. Conversely, let z ≤ a∨ bw
for
all w , we prove z ≤ a∨ inf w
b w
.
Case 1: Let a ≤ inf w
b w
. Then z ≤ bw
for each w ,
z ≤ inf w
b w
and z ≤ a∨ inf w
b w
.
Case 2: Let a ≥ inf w
b w
. Then for some w 0
, a ≥ b w0
,
thus z ≤ a and z ≤ a∨ inf w
b w
.
So we prove the soundness of axioms.
Theorem 6 (Soundness of deduction rules) (1) For
arbitrary formulas φ, ψ , safe-structure M and evaluation
L L L
v , ψ ≥ φ ∗ φ → ψ . In particular, if
M, v M, v M,
v
L
L
L
= → = 1
M, v
M,
v L
then ψ 1
M ,
=
v L
.
L L L
(2) Consequently, ψ φ φ ψ
M M M
φ φ ψ
≥ ∗ → , thus if
φ, φ → ψ are then ψ is 1 L
-true in M.
(3) If φ is 1 L
-true in M then Δ φ is 1 L
-true in M .
(4) If φ is 1 L
-true in M then ( ∀ x)
φ is 1 L
-true in M.
Proof. (1) is just as in propositional calculus.
To prove (2) put φ = aw, ψ = bw, inf
w
w
w
aw
= a. We
have to prove inf
w( aw ⇒bw) ≤infw aw ⇒ infwbw
.
Observe the following:
inf( aw ⇒bw) ≤( aw ⇒bw) ≤( a⇒ bw)
,
thus inf
w( aw ⇒bw) ≤inf w( a⇒ bw)
. It remains to prove
inf
w( a⇒bw) ≤ a⇒ infwbw, this is holds from Theorem
5.
L
(3) If φ is 1 L
-true in M then φ
M
= 1L
, so
L
L
Δ φ =Δ φ = . Then (3) holds.
1
M, v
M,
v L
L
M
φ ′
′
,
}
L
M
L
M
(4) Being φ = inf{ φ ,v
| v M − evaluation}
≤ inf{
Mv| v ≡ v = ( ∀ x) φ } , So (4) holds.
So we can get the following soundness theorem.
L
Theorem 7 (Soundness) Let L is linearly ordered
ŁΠG − -algebra and φ is a formula in J, if φ , then φ is
L
L-tautology, i.e. φ = 1 . M L
Similarly, we can get the following strong soundness
theorem.
Definition 10 Let T be a theory, L be a linearly ordered
ŁΠG − -algebra and M a safe L-structure for the language
of T. M is an L-model of T if all axioms of T are 1 L
-true
in M, i.e. φ = 1 L
in each φ ∈ T .
Theorem 8 (Strong Soundness) Let T be a theory, L is
linearly ordered ŁΠG − -algebra and φ is a formula in J,
L
if T φ ( φ is provable in T), then φ
M
= 1L
for each
linearly ordered ŁΠG − -algebra L and each L-model M
of T.
Proof. In fact, from the proof of Theorem 5, for each
L-model M of T, the axioms are true, and the formulas in
T are true, from the proof of Theorem 6, the deduction
rules preserve true. So the theorem holds.
Theorem 9 (Deduction Theorem) Let T be a theory,
φ, ψ are closed formulas. Then ( T ∪φ)
ψ iff
T Δφ →ψ
.
Proof. Sufficiency: Let T Δφ →ψ
, from φ
( φ∈ ( T ∪ φ)
), then Δ φ by necessitation, so we can get
ψ by MP rules.
Necessity: Let m is the proof length from T ∪ φ}
to ψ ,
we prove by induction for the length m.
When m = 1 , ψ ∈T
∪φ∪Axm(C ∀ ) , if ψ = φ , The
result holds. If ψ ∈ T or ψ is axiom, from Lemma2 (2),
we have ψ →( Δφ → ψ)
, then by ψψ , →( Δφ→ ψ)
, we
get Δφ → ψ , thus T Δφ →ψ
.
Assume that the result holds when m≤
k , i.e. we get
γ at k step, then T Δφ → γ . Now Let m= k + 1.
If ψ is obtained from MP rule by the above results
γγ , → ψ in the proof sequence, then by induction
hypothesis, we get T Δφ → γ, T Δφ →( γ →ψ)
.
From Lemma 3, we can get
T ( Δφ& Δφ) →( γ &( γ →ψ))
. Being
T ( Δφ) &( Δφ)
≡ Δφ
, so T Δφ →( γ &( γ →ψ))
.
From lemma 2 (4) we have ( γ & ( γ →ψ)
→ ψ , so we
get T Δφ →ψ
by the hypothetical syllogism.
If ψ is obtained from necessitation rule by the above
step γ in the proof sequence, i.e. Δ γ = ψ , then by
induction hypothesis, we get T Δφ →γ
.
T Δ( Δφ →γ)
, from (U16) we can get T ΔΔφ → Δγ
,
from (U15) we can get Δφ
→ΔΔ φ , thus by the
hypothetical syllogism we can get T Δφ →Δγ
, i.e.
T Δφ →ψ
.
If ψ is obtained from generalization rule by the above
step γ in the proof sequence, i.e. ( ∀ x)
γ = ψ , then by
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