In this section, we introduce and characterize complete decomposable *MS*-triples of complete decomposable *MS*-algebras.

Let *L* be a decomposable *MS*-algebra *L*. For *ϕ*≠*N*⊆*L*, define *N*^{∘} as follows:

*N*^{∘}={*n*^{∘}:*n*∈*N*}.

###
**Lemma 2**

If *L* is a complete decomposable *MS*-algebra, then for *ϕ*≠*N*⊆*L*,*ϕ*≠*C*⊆*L*^{∘∘} and *ϕ*≠*E*⊆*D*(*L*), we have (1) (sup*L**N*)^{∘}= inf*L**N*^{∘}, (2) \(\sup _{L^{\circ \circ }}C=(\sup _{L}C)^{\circ \circ }=(\inf _{L} C^{\circ })^{\circ }\), (3) \(\inf _{L^{\circ \circ }} C=\inf _{L}C\), (4) inf*D*(*L*)*E*= inf*L**E* and sup*D*(*L*)*E*= sup*L**E*.

###
*Proof*

(1). Let *x*= sup*L**N*. Then *x*≥*n* for all *n*∈*N* implies *x*^{∘}≤*n*^{∘}. Hence *x*^{∘} is a lower bound of *N*^{∘}. Let *y* be a lower bound of *N*^{∘}. Then *y*≤*n*^{∘} for all *n*∈*N* implies *y*^{∘}≥*n*^{∘∘}≥*n*. So, *y*^{∘} is an upper bound of *N*. Thus *x*≤*y*^{∘} as *x*= sup*L**N*. This gives *x*^{∘}≥*y*^{∘∘}≥*y*. Therefore *x*^{∘}= inf*L**N*^{∘}=(sup*L**N*)^{∘}. (2) Let sup*L**C*=*x*. Then *x*^{∘∘}=(sup*L**C*)^{∘∘}. We have to show that \(x^{\circ \circ }=\sup _{L^{\circ \circ }}C\). Since sup*L**C*=*x*, then *x*≥*c* for all *c*∈*C*. so, *x*^{∘∘}≥*c*^{∘∘}=*c* for all *c*∈*C*. Therefore *x*^{∘∘} is an upper bound of *C*. Let *y* be another upper bound of *C* in *L*^{∘∘}. Then *y*≥*c* for all *c*∈*C*. Thus *y*^{∘∘}≥*c*^{∘∘}=*c*. Hence *y*^{∘∘} is an upper bound of *C*. So *y*^{∘∘}≥*x* as *x*= sup*L**C*. It follows that *y*=*y*^{∘∘}≥*x*^{∘∘}. Hence *x*^{∘∘} is the least upper bound of *C*. Since *x*^{∘∘}∈*L*^{∘∘}, then \(x^{\circ \circ }=\sup _{L^{\circ \circ }}C\). By (1) we have (sup*L**C*)^{∘∘}=(inf*L**C*^{∘})^{∘}. (3) Let *x*= inf*L**C*. Then *x*≤*c* for all *c*∈*C*. Then *x*^{∘∘}≤*c*^{∘∘}=*c*. Hence *x*^{∘∘} is a lower bound of *C*. Thus *x*≥*x*^{∘∘} as *x*= inf*L**C*. But *x*≤*x*^{∘∘}. Then *x*^{∘∘}=*x* and *x*∈*L*^{∘∘}. Thus \(\inf _{L^{\circ \circ }}C=x\). (4) Let *x*= inf*L**E* and *y*= inf*D*(*L*)*E*. Then *x*≤*e* and *y*≤*e* for all *e*∈*E* imply that *x*=*y*. Now we prove sup*D*(*L*)*E*= sup*L**E*. Let *y*= sup*L**E*. Then *y*≥*e* for all *e*∈*E*. It follows that *y*^{∘}≤*e*^{∘}=0. Then *y*∈*D*(*L*) implies *y*= sup*D*(*L*)*E*. □

Let (*M*,*D*,*φ*) be a decomposable *MS*-triple. For any *∅*≠*E*⊆*D*, consider the set *M*_{E} as follows:

\(M_{E}=\left \{a\in M : \bar {a}\varphi \vee [z)\supset E \ \text {for some} \ z\in D\right \}\).

###
**Lemma 3**

Let (*M*,*D*,*φ*) be a decomposable MS-triple. For any *∅*≠*E*⊆*D*, we have (1) *M*_{E} is an ideal of *M*, (2) [*E*)=∪{[*t*):*t*∈*E*}, (2) *M*_{E}=*M*_{[E)}.

###
*Proof*

(1). Let *a*,*b*∈*M*_{E}. Then \(\bar {a}\varphi \vee [z_{1})\supset E\) and \(\bar {b}\varphi \vee [z_{2})\supset E\) for some *z*_{1},*z*_{2}∈*D*. Hence \(E\subset (\bar {a}\varphi \vee [z_{1})) \cap (\bar {b}\varphi \vee [z_{2}))=\overline {(a\vee b)}\varphi \vee [t)\) for some *t*∈*D* (see Theorem 2). It follows that *a*∨*b*∈*M*_{E}. Now, let *a*∈*M*_{E} and *c*∈*M*. Then, ∃*z*∈*D* such that \(\bar {a}\varphi \vee [z)\supset E\). Since *a*∧*c*≤*a*, then \(\overline {a\wedge c}\geq \bar {a}\). This gives \(\overline {(a\wedge c)}\varphi \supseteq \bar {a}\varphi \). It follows that \(\overline {(a\wedge c)}\varphi \vee [z)\supseteq \bar {a}\varphi \vee [z)\supset E\). Then *a*∧*c*∈*M*_{E}. Consequently, *M*_{E} is an ideal of *M*. (2) Obvious. (3) Clearly, *M*_{[E)}⊆*M*_{E}. Let *a*∈*M*_{E}. Then, ∃*z*∈*D* such that \(\bar {a}\varphi \vee [z)\supset E\). Since \(\bar {a}\varphi \vee [z)\) is a filter of *D* and [*E*) is the smallest filter of *D* containing *E*, then \(\bar {a}\varphi \vee [z)\supset [E)\). Hence, *a*∈*M*_{[E)} and *M*_{E}⊆*M*_{[E)}. Therefore, *M*_{E}=*M*_{[E)}. □

###
**Definition 8**

A complete decomposable *MS*-triple is a decomposable *MS*-triple (*M*,*D*,*φ*) satisfying the following conditions: (i) *M* is complete, (ii) *D* is conditionally complete, (iii) For each *∅*≠*E*⊆*D*, the set *M*_{E} has the greatest element in *M*.

###
**Theorem 3**

Let *L* be a complete decomposable *MS*-algebra constructed from the decomposable *MS*-triple (*M*,*D*,*φ*). Then, the triple (*M*,*D*,*φ*) is complete.

###
*Proof*

Since *L* is associated with the decomposable *MS*-triple (*M*,*D*,*φ*), then by Theorem 2, we have

\(L=\left \{(a,\bar {a}\varphi \vee [x)): a\in M,x\in D\right \}\).

Corollary 1(1)-(3), gives

\(L^{\circ \circ }=\left \{(a,\bar {a}\varphi): a\in M\right \}\cong M\) and *D*(*L*)={(1,[*x*)):*x*∈*D*}≅*D*.

We have to prove that a decomposable *MS*-triple (*M*,*D*,*φ*) is complete. So we proceed to prove (i)–(iii) of Definition 8. For (i), let *∅*≠*C*⊆*M*. Consider a subset \(\acute {C}=\{(c,\bar {c}\varphi):c\in C\}\) of *L*^{∘∘} corresponding to *C*. Since *L* is complete, then \(\inf _{L}\acute {C}=(a,\bar {a}\varphi \vee [x))\) for some \((a,\bar {a}\varphi \vee [x))\in L\). Thus, \((a,\bar {a}\varphi \vee [x))\leq (c,c\varphi)\) for all *c*∈*C*. Then *a*≤*c* for all *c*∈*C* implies that *a* is a lower bound of *C*. We verify that *a* is the greatest lower bound of *C* in *M*. Let *b* be a lower bound of *C*. Then *b*≤*c* for all *c*∈*C*. This gives \(\bar {b}\varphi \supseteq \bar {c}\varphi \). Therefore, \((b,\bar {b}\varphi)\leq (c,\bar {c}\varphi)\) for all *c*∈*C* and (*b*,*b**φ*) is a lower bound of \(\acute {C}\). Then \((a,\bar {a}\varphi \vee [x))\geq (b,b\varphi)\) as \(\inf _{L}C=(a,\bar {a}\varphi \vee [x))\). Consequently, *a*≥*b* and *a*= inf*M**C*. Since *a*= inf*M**C* and *M* is bounded above by 1, then, *M* is complete.Now we prove (ii). Let *ϕ*≠*E*⊆*D*. Consider \(\acute {E}\subseteq D(L)\) corresponding to *E*. Then

\(\acute {E}=\left \{(1,[e)): e\in D\right \}\).

Let *z* be a lower bound of *E*. Since *L* is complete, then \(\inf _{L}\acute {E}\) exists. Let \(\inf _{L}\acute {E}=(a,\bar {a}\varphi \vee [x))\). Since *z*≤*e* for all *e*∈*E* as *z* is a lower bound of *E*. Then, [*z*)⊇[*e*) and (1,[*z*))≤(1,[*e*)). Thus, (1,*z*) is a lower bound of \(\acute {E}\). Then, \((a,\bar {a}\varphi \vee [x))\geq (1,[z))\) because of \(\inf _{L}\acute {E}=(a,\bar {a}\varphi \vee [x))\). This implies that *a*≥1 and \(\bar {a}\varphi \vee [x)\subseteq [z)\). Consequently, *a*=1 and \(\bar {a}\varphi \vee [x)=0\varphi \vee [x)=[x)\). Thus [*x*)⊆[*z*) implies *x*≥*z*. This shows that *x* is the greatest lower bound of *E* in *D* and *x*= inf*D**E*. Using a similar way, we can show that, if *E* has an upper bound, then sup*D**E* exists. Therefore, *D* is a conditionally complete lattice as required.

Now we prove (iii). Let *∅*≠*E*⊆*D*. Consider \(\acute {E}\subseteq D(L)\) corresponding to *E*. Then

$$\begin{array}{@{}rcl@{}} \acute{E}=\left\{(1,[x)): x\in E\right\}. \end{array} $$

Since *L* is complete, then \(\inf _{L} \acute {E}\) exists. Let \((b,\bar {b}\varphi \vee [z))=\inf _{L} \acute {E}\). We show that *b* is the largest element of *M*_{E}. Since \((b,\bar {b}\varphi \vee [z))=\inf _{L} \acute {E}\), then \((b,\bar {b}\varphi \vee [z))\leq (1,[x)), \ \forall x\in E\). This gives *b*≤1 and \(\bar {b}\varphi \vee [z)\supseteq [x), \ \forall x\in E\). Therefore, \(\bar {b}\varphi \vee [z)\supseteq \cup _{x\in E}[x)=[E)\supset E\). Thus, *b*∈*M*_{E}. Now, let *c*∈*M*_{E}. Then \(\bar {c}\varphi \vee [y)\supset E\) for some *y*∈*D*. It follows that \(\bar {c}\varphi \vee [y)\supseteq [E)\supseteq [x)\) for all *x*∈*E*. Hence, \((1,[x))\leq (c,\bar {c}\varphi \vee [y))\) for all *x*∈*E*. Thus, \((c,\bar {c}\varphi \vee [y))\) is a lower bound of \(\acute {E}\) and therefore \((c,\bar {c}\varphi \vee [y))\leq (b,\bar {b}\varphi \vee [z))\). Then, *c*≤*b*. This deduce that *b* is the largest element of *M*_{E} in *M*. Therefore, (*M*,*D*,*φ*) is a complete decomposable *MS*-triple. □

The converse of the above theorem is given in the following.

###
**Theorem 4**

Let *L* be a decomposable *MS*-algebra constructed from the complete decomposable *MS*-triple (*M*,*D*,*φ*). Then *L* is complete.

###
*Proof*

Let (*M*,*D*,*φ*) be a complete decomposable *MS*-triple. Then –(iii) of Definition 8 hold. Let *∅*≠*N*⊆*L*, where *L* is constructed as in construction Theorem from the decomposable *MS*-triple (*M*,*D*,*φ*) as follows:

\(L=\left \{(a,\bar {a}\varphi \vee [x)):a\in M,x\in D\right \}\).

Since *L* is bounded, it is enough to show the existence of inf*L**N*. Denote *a*= inf*M**N*^{∘∘} and \(F=\cup \left \{[t): (c,\bar {c}\varphi \vee [t))\in N \text {for some} \ c\in M\right \}\) (∪ means the union in *F*(*D*)). Let *b*= max*M*_{F}. Now, we prove that there exists an element *z*∈*D* such that \(\bar {b}\varphi \vee [z)\supset F\) and if \(\bar {b}\varphi \vee [y)\supset F\) for some *y*∈*D* then \(\bar {b}\varphi \vee [y) \supseteq \bar {b}\varphi \vee [z)\). For this purpose, consider the following set:

\(\left \{x_{\gamma }:\gamma \in \Gamma \text {for all}x_{\gamma }\text {with}\bar {b}\varphi \vee [x_{\gamma })\supset F\right \}\).

Thus, we have to find a *z*∈*D* with \(\bar {b}\varphi \vee [y)\supset F\) and \(\bar {b}\varphi \vee [y)\supseteq \bar {b}\varphi \vee [z)\) for all *γ*∈*Γ*. The set \(\left \{x_{\gamma }:\gamma \in \Gamma \text {for all} x_{\gamma }\text {with}\bar {b}\varphi \vee [x_{\gamma })\supset F\right \}\) is bounded from above. Then, by (ii), there exists *s*= sup*D*{*x*_{γ}:*γ*∈*Γ*}. We prove that ∩_{γ∈Γ}[*x*_{γ})=[*s*).

$$\begin{array}{@{}rcl@{}} y\in\cap_{\gamma\in\Gamma}[x_{\gamma})&\Leftrightarrow& y\in[x_{\gamma}),~ \ \forall\gamma\in\Gamma\\&\Leftrightarrow& y\geq x_{\gamma},~ \ \forall\gamma\in\Gamma\\ &\Leftrightarrow& y \ \text{is an upper bound of} \ \{x_{\gamma}: \gamma\in\Gamma\}\\ &\Leftrightarrow& y\geq s\ \text{as}\ s=\sup_{D}\{x_{\gamma} : \gamma\in\Gamma\}\\&\Leftrightarrow& y\in[s). \end{array} $$

Then it is sufficient to prove the following equality.

$$\begin{array}{@{}rcl@{}} \cap_{\gamma\in\Gamma}(\bar{b}\varphi\vee[x_{\gamma}))=\bar{b}\varphi\vee\cap_{\gamma\in\Gamma}[x_{\gamma})=\bar{b}\varphi\vee[s). \end{array} $$

(1)

Let \(t\in \bar {b}\varphi \vee [s)\). Then

$$\begin{array}{@{}rcl@{}} t\in \bar{b}\varphi\vee[s)&\Rightarrow& t\geq t_{1}\wedge s\ \text{where}\ t_{1}\in \bar{b}\varphi\\ &\Rightarrow& t\geq t_{1}\wedge (s\vee x_{\gamma})\ \text{as}\ s\geq x_{\gamma}\text{for all}\ \gamma\in \Gamma\\ &\Rightarrow&t\geq(t_{1}\wedge s)\vee (t_{1}\wedge x_{\gamma})\\ &\Rightarrow&t\geq t_{1}\wedge x_{\gamma}\\ &\Rightarrow& t\in \bar{b}\varphi\vee[x_{\gamma})\ \text{for all}\ \gamma\in \Gamma. \end{array} $$

Then \(\bar {b}\varphi \vee \cap _{\gamma \in \Gamma }[x_{\gamma })\subseteq \bar {b}\varphi \vee [x_{\gamma })\) implies \(\bar {b}\varphi \vee \cap _{\gamma \in \Gamma }[x_{\gamma })\subseteq \cap _{\gamma \in \Gamma }(\bar {b}\varphi \vee [x_{\gamma }))\). Conversely, let \(y\in \cap _{\gamma \in \Gamma }(\bar {b}\varphi \vee [x_{\gamma }))\). Then \(y\in \bar {b}\varphi \vee [x_{\gamma })\) for all *γ*∈*Γ*. Hence *y*≥*t*∧*z* for \(t\in \bar {b}\varphi \) and *z*∈[*x*_{γ}) for all *γ*∈*Γ*. It follows that *z*≥*x*_{γ} for all *γ*∈*Γ*. This means that *z* is an upper bound of the set {*x*_{γ}:*γ*∈*Γ*}. Then *s*≤*z* as *s*= sup*D*{*x*_{γ}:*γ*∈*Γ*}. Now

$$\begin{array}{@{}rcl@{}} y&\geq& t\wedge z\\ &=&t\wedge(s\vee z)\ \text{as}\ s\leq z\\ &=&(t\wedge s)\vee(t\wedge z)\ \text{by distributivity of}\ D\\ &\geq&t\wedge s\in \bar{b}\varphi\vee[s). \end{array} $$

Then \(y\in \bar {b}\varphi \vee [s)\). Therefore, \(\cap _{\gamma \in \Gamma }(\bar {b}\varphi \vee [x_{\gamma }))\subseteq \bar {b}\varphi \vee [s)\).

We prove the existence of inf*L**N*. First, we claim that

\(i=\left (a\wedge b,\overline {(a\wedge b)}\varphi \vee [z)\right)=\inf _{L} N~(\text {we put then} z=s)\).

First, we show that *i* is a lower bound of *N*. Let \((f,\bar {f}\varphi \vee [y))\in N.\) Since *a*= inf*M**N*^{∘∘}, we get *a*≤*f*. So, *a*∧*b*≤*a*≤*f*. Then *a*∧*b*≤*f* implies that \(\overline {a\wedge b}\geq \bar {f}\). Consequently, \(\overline {(a\wedge b)}\varphi =\bar {a}\varphi \vee \bar {b}\varphi \supseteq \bar {f}\varphi \). Moreover, \([y)\subseteq F\subseteq \bar {b}\varphi \vee [z)\) as *y*∈*F*. Then

$$\begin{array}{@{}rcl@{}} \overline{(a\wedge b)}\varphi\vee[z)&=&(\bar{a}\vee \bar{b})\varphi\vee[z)\\ &=&(\bar{a}\varphi\vee \bar{b}\varphi)\vee(\bar{b}\varphi\vee[z))\\ &\supseteq&\bar{f}\varphi\vee[y). \end{array} $$

Then \((a\wedge b,\overline {(a\wedge b)}\varphi \vee [z))\leq (f,\bar {f}\vee [y))\) for all \((f,\bar {f}\vee [y))\in N\). Therefore, *i* is a lower bound of *N*. It remains to show that *i* is the greatest lower bound of *N*. Let \((c,\bar {c}\varphi \vee [x))\) be a lower bound of *N*. Then, \((c,\bar {c}\varphi \vee [x))\leq (f,\bar {f}\varphi \vee [y)), \ \forall (f,\bar {f}\varphi \vee [y))\in N\). So, *c*≤*f*, ∀*f*∈*N*^{∘∘}. Then *c* is a lower bound of *N*^{∘∘}. Thus *c*≤*a* as \(a=\inf \limits _{M} N^{\circ \circ }\). On the other hand, \(\bar {c}\varphi \vee [x)\supseteq \bar {f}\varphi \vee [y), \ \forall (f,\bar {f}\varphi \vee [y))\in N\). So, \(\bar {c}\varphi \vee [x)\supseteq [y), \ \forall y\in F\). Therefore, \(\bar {c}\varphi \vee [x)\supseteq F\). Hence, \(\bar {c}\varphi \vee [x)\supseteq \bar {b}\varphi \vee [z)\) by using equality (1). Then \(\bar {c}\varphi \vee [x)\supseteq F\) implies that *c*∈*M*_{F}. So, *c*≤*b* as \(b=\max \limits _{M} M_{F}\in M\). Now, we have *c*≤*a* and *c*≤*b*. Then *c*≤*a*∧*b*. Moreover, we have \(\bar {c}\varphi \supseteq \bar {a}\varphi \) because of *c*≤*a*. Also, \(\bar {c}\varphi \vee [x)\supseteq \bar {b}\varphi \vee [z)\). So, \(\bar {c}\varphi \vee [x)\supseteq \bar {a}\varphi \vee \bar {b}\varphi \vee [z)=\overline {(a\wedge b)}\varphi \vee [z)\). Therefore, \((c,\bar {c}\varphi \vee [x))\leq i\). Then *i*= inf*L**N* and *L* is complete. □

###
**Corollary 2**

If *M* and *D* are complete, then so is *L*.

###
*Proof*

. We need only to prove that the condition (*i**i**i*) of Definition 8 holds. Let *E*⊆*D* and *t*= inf*D**E*. Then, [*t*)=[ inf*D**E*)⊇*E*. So, \((1,\bar {1}\varphi \vee [t))=(1,[t))\in L\). Therefore, 1∈*M*_{E}. Hence, by the above Theorem, *L* is complete. □

###
**Corollary 3**

If *M* is finite and *D* is conditionally complete, then *L* is complete.

###
*Proof*

Since *M* is finite and *M*_{E} is an ideal of *M* (see Lemma 1(1)), then *M* is complete and *M*_{E} is a principal ideal of *M*. Therefore, *M*_{E} contains the greatest element in *M*. So, the conditions (*i*)– (*i**i**i*) of Definition 8 are satisfied and consequently, *L* is complete. □

Combining Theorems 3 and 4, we get the following theorem.

###
**Theorem 5**

Let *L* be a decomposable *MS*-algebra constructed from the decomposable *MS*-triple (*M*,*D*,*φ*). Then *L* is complete if and only if (*M*,*D*,*φ*) is complete.

Let *L* be a complete decomposable *MS*-algebra. In the proof of Theorem 4 arbitrary meets in *L* are described. In the following Lemma, we describe joins in *L*.

###
**Lemma 4**

Let *L* be a complete decomposable *MS*-algebra constructed from the decomposable *MS*-triple (*M*,*D*,*φ*). Let *ϕ*≠*N*⊆*L* and *a*= sup*M**N*^{∘∘}. Then there exists an element *z*∈*D* such that \([z)=\bigcap \left \{\bar {c}\varphi \vee [t):(c,\bar {c}\varphi \vee [t))\in N\right \}\cap a\varphi \) and \(\sup N=(a,\bar {a}\varphi \vee [z))\).

###
*Proof*

Let *ϕ*≠*N*⊆*L* and \(\sup _{L} N=(b,\bar {b}\varphi \vee [z))\). We can assume that *z*∈*a**φ*. We prove that *b*=*a*= sup*M**N*^{∘∘}. Using Lemma 2(2), we get

\(\sup _{M}N^{\circ \circ }=(\sup _{L}N)^{\circ \circ }=(b,\bar {b}\varphi \vee [z))^{\circ \circ }=(b,\bar {b}\varphi)\).

But \(a=(a,\bar {a}\varphi)=\sup _{M}N^{\circ \circ }\). Then *b*=*a*. Hence, \(\bar {a}\varphi \vee [z)\) is the greatest filter of the form \(\bar {a}\varphi \vee [x), x\in D\) with

\(\bar {a}\varphi \vee [z))\subset \bar {c}\varphi \vee [t)\) for each \((c,\bar {c}\varphi \vee [t))\in N\).

The last condition is equivalent to

\([z)\subset \bigcap \left \{\bar {c}\varphi \vee [t):(c,\bar {c}\varphi \vee [t))\in N\right \}\cap a\varphi \).

Let \( \bigcap \left \{\bar {c}\varphi \vee [t):(c,\bar {c}\varphi \vee [t))\in N\right \}\cap a\varphi =R\). If [*z*)≠*R*, then there is *y*∈*R*,*y*≧̸*z*. It follows that *y*∧*z*<*z* and *y*∧*z*∈*R*. Then [*z*)⊂[*y*∧*z*) implies \(\bar {a}\varphi \vee [z)\subset \bar {a}\varphi \vee [y\wedge z)\). Since *y*∧*z*∈*R* then \([y\wedge z)\subset \bar {c}\varphi \vee [t)\) for all \((c,\bar {c}\varphi \vee [t))\in N\). Since *a*≥*c* (as *a*= sup*M**N*^{∘∘}) then \(\bar {a}\leq \bar {c}\). It follows that \(\bar {a}\varphi \leq \bar {c}\varphi \). Therefore, \(\bar {a}\varphi \vee [y\wedge z)\subset \bar {c}\varphi \vee [t)\) for all \((c,\bar {c}\varphi \vee [t))\in N\). Consequently,

\(\bar {a}\varphi \vee [z)\subset \bar {a}\varphi \vee [y\wedge z)\subset \bar {c}\varphi \vee [t)\) for all \((c,\bar {c}\varphi \vee [t))\in N\),

which contradicts the maximality of \(\bar {a}\varphi \vee [z)\). □