Three-valued semantic pluralism: a defense of a three-valued solution to the sorites paradox

Abstract

Disagreeing with most authors on vagueness, the author proposes a solution that he calls ‘three-valued semantic pluralism’ to the age-old sorites paradox. In essence, it is a three-valued semantics for a first-order vague language with identity with the additional suggestion that a vague language has more than one correct interpretation. Unlike the traditional three-valued approach to a vague language, three-valued semantic pluralism can accommodate the phenomenon of higher-order vagueness and the phenomenon of penumbral connection when equipped with ‘suitable conditionals’. The author also shows that three-valued semantic pluralism is a natural consequence of a restricted form of the Tolerance principle ($\hbox {T}_R$) and a few related ideas, and argues that ($\hbox {T}_R$) is well-motivated by considerations about how we learn, teach, and use vague predicates.

Appendix

Our goal in this appendix is to prove Theorem 1, i.e., that a disjunction is true simpliciter iff at least one of its disjuncts is true simpliciter. Throughout this appendix, ‘($\forall {\mathfrak {M}} \in {\mathbb {S}}$)...’ is short for ‘$\forall {\mathfrak {M}}$(${\mathfrak {M}} \in {\mathbb {S}} \rightarrow \ldots$)’ while ‘($\exists {\mathfrak {M}} \in {\mathbb {S}}$)...’ is short for ‘$\exists {\mathfrak {M}}$(${\mathfrak {M}} \in {\mathbb {S}} \wedge \ldots$)’. I first restate here two assumptions that I made in Sect. 3:

Assumption (A)

($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($\forall {\mathfrak {M'}} \in {\mathbb {S}}$)(($v_{\mathfrak {M}}$(p) = $n\,\wedge \,v_{\mathfrak {M'}}$(q) = n) $\rightarrow$ ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}}\,\le {\mathfrak {M}}\,\wedge \,\mathfrak {M^{*}}\,\le {\mathfrak {M'}}\,\wedge \,v_\mathfrak {M^{*}}$(p) = $v_\mathfrak {M^{*}}$(q) = n)), for any atomic sentences p and q.

Assumption (B)

(($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(p) $\ne 1$) $\wedge$ ($\exists {\mathfrak {M'}} \in {\mathbb {S}}$)($v_{\mathfrak {M'}}$(p) $\ne 0$)) $\rightarrow$ ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$) ($v_\mathfrak {M^{*}}$(p) $= n$), for any atomic sentence p.

With these two assumptions, we will prove, as lemmas of Theorem 1, that Assumption (A) and Assumption (B) are not only true of atomic sentences but also true of complex ones. Before giving the proofs, however, we state, without giving the proof, a very famous result about strong $\hbox {K}_3$ and many other three-valued semantics, i.e., Proposition 1. In Proposition 1, the relation $\le$ between truth-values is defined as: $n \le n, 0 \le 0, 1 \le 1, n \le 0$, and $n \le 1$.

Proposition 1

If ${\mathfrak {M}} \le {\mathfrak {M'}}$, then $v_{\mathfrak {M}}$(A) $\le v_{\mathfrak {M'}}$(A), for every sentence A of ${\mathfrak {L}}$.

We now set out our task. We will prove that Assumption (A) can be generalized to all sentences, i.e., Lemma 2. As a midway to Lemma 2, we prove Lemma 1 first.

Lemma 1

Let p be any atomic sentence and B be any sentence. Then ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($\forall {\mathfrak {M'}} \in {\mathbb {S}}$)(($v_{\mathfrak {M}}$(p) $= n\,\wedge \,v_{\mathfrak {M'}}$(B) $= n$) $\rightarrow$ ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}} \le {\mathfrak {M}} \wedge \mathfrak {M^{*}} \le {\mathfrak {M'}}\,\wedge \,v_\mathfrak {M^{*}}$(p) $= v_\mathfrak {M^{*}}$(B) $= n$)).

Proof

We prove Lemma 1 by induction on the number of connectives in B. For simplicity, we assume that ‘$\wedge$’ is defined in terms of ‘$\lnot$’ and ‘$\vee$’ and we omit the quantificational case.

• Base case This is automatically true by Assumption (A).

• Inductive step Assume that Lemma 1 is true of sentences C and D whose numbers of connectives are less than that in B. Two cases:

• Cases 1 B is ‘$\lnot$C’

  Assume that $v_{\mathfrak {M}}$(p) $= n$ and $v_{\mathfrak {M'}}$(B) = $v_{\mathfrak {M'}}$($\lnot$C) $= n$ for some $\mathfrak {M, M'} \in {\mathbb {S}}$. Then $v_{\mathfrak {M'}}$(C) $= n$. By inductive hypothesis, ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}} \le {\mathfrak {M}} \wedge \mathfrak {M^{*}} \le {\mathfrak {M'}} \wedge v_\mathfrak {M^{*}}$(p) $= v_\mathfrak {M^{*}}$(C) $= n$). But then $(\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}} \le {\mathfrak {M}} \wedge \mathfrak {M^{*}}\,\le {\mathfrak {M'}} \wedge v_\mathfrak {M^{*}}$(p) $= v_\mathfrak {M^{*}}$(C) $= v_\mathfrak {M^{*}}$($\lnot$C) $= v_\mathfrak {M^{*}}$(B) $= n$).

• Case 2 B is ‘C $\vee$ D’

  Assume that $v_{\mathfrak {M}}$(p) $= n$ and $v_{\mathfrak {M'}}$(B) $= v_{\mathfrak {M'}}$(C $\vee$ D) $= n$ for some $\mathfrak {M, M'} \in {\mathbb {S}}$. There are three sub-cases. In each sub-case, we prove that $(\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}} \le {\mathfrak {M}} \wedge \mathfrak {M^{*}} \le {\mathfrak {M'}} \wedge v_\mathfrak {M^{*}}$(p) $= v_\mathfrak {M^{*}}$(C $\vee$ D) $= v_\mathfrak {M^{*}}$(B) $= n$).

• Sub-case 2a$v_{\mathfrak {M'}}$(C) $= n$ but $v_{\mathfrak {M'}}$(D) $= 0$.

  In this case, $v_{\mathfrak {M}}$(p) $= n$ and $v_{\mathfrak {M'}}$(C) $= n$. So, by inductive hypothesis, ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}} \le {\mathfrak {M}} \wedge \mathfrak {M^{*}} \le {\mathfrak {M'}} \wedge v_\mathfrak {M^{*}}$(p) $= v_\mathfrak {M^{*}}$(C) $= n$). Since $\mathfrak {M^{*}} \le {\mathfrak {M'}}$, it follows that $v_\mathfrak {M^{*}}$(D) $\le v_{\mathfrak {M'}}$(D) by Proposition 1. Since $v_{\mathfrak {M'}}$(D) $= 0$ and $v_\mathfrak {M^{*}}$(D) $\le v_{\mathfrak {M'}}$(D), it further follows that $v_\mathfrak {M^{*}}$(D) $= 0$ or $v_\mathfrak {M^{*}}$(D) $= n$. Either way, ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}} \le {\mathfrak {M}} \wedge \mathfrak {M^{*}} \le {\mathfrak {M'}} \wedge v_\mathfrak {M^{*}}$(p) $= v_\mathfrak {M^{*}}$(C $\vee$ D) $= v_\mathfrak {M^{*}}$(B) $= n$).

• Sub-case 2b$v_{\mathfrak {M'}}$(C) $= 0$ but $v_{\mathfrak {M'}}$(D) $= n$.

  The proof of sub-case 2b is similar to that of sub-case 2a.

• Sub-case 2c$v_{\mathfrak {M'}}$(C) $= n$ but $v_{\mathfrak {M'}}$(D) $= n$.

  The proof of sub-case 2c is similar to that of sub-case 2a. $\square$

Lemma 2

($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($\forall {\mathfrak {M'}} \in {\mathbb {S}}$)(($v_{\mathfrak {M}}$(A) $= n \wedge v_{\mathfrak {M'}}$(B) $= n$) $\rightarrow$ ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}} \le {\mathfrak {M}} \wedge \mathfrak {M^{*}} \le {\mathfrak {M'}} \wedge v_\mathfrak {M^{*}}$(A) $= v_\mathfrak {M^{*}}$(B) $= n$)), for any sentences A and B.

Proof

We prove Lemma 2 by induction on the number of connectives in A. Again for simplicity, we assume that ‘$\wedge$’ is defined in terms of ‘$\lnot$’ and ‘$\vee$’ and we omit the quantificational case.

• Base case This is automatically true by Lemma 1.

• Inductive step Assume that Lemma 2 is true of sentences C and D whose numbers of connectives are less than that in A. Two cases:

• Cases 1 A is ‘$\lnot$C’

  Assume that $v_{\mathfrak {M}}$(A) $= v_{\mathfrak {M}}$($\lnot$C) $= n$ and $v_{\mathfrak {M'}}$(B) $= n$ for some $\mathfrak {M, M'} \in {\mathbb {S}}$. Then $v_{\mathfrak {M}}$(C) $= n$. By inductive hypothesis, ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}} \le {\mathfrak {M}} \wedge \mathfrak {M^{*}} \le {\mathfrak {M'}} \wedge v_\mathfrak {M^{*}}$(C) $= v_\mathfrak {M^{*}}$(B) $= n$). But then ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}} \le {\mathfrak {M}} \wedge \mathfrak {M^{*}} \le {\mathfrak {M'}} \wedge v_\mathfrak {M^{*}}$(C) $= v_\mathfrak {M^{*}}$($\lnot$C) $= v_\mathfrak {M^{*}}$(A) $= v_\mathfrak {M^{*}}$(B) $= n$).

• Case 2 A is ‘C $\vee$ D’

  Assume that $v_{\mathfrak {M}}$(A) $= v_{\mathfrak {M}}$(C $\vee$ D) $= n$ and $v_{\mathfrak {M'}}$(B) $= n$ for some $\mathfrak {M, M'} \in {\mathbb {S}}$. There are three sub-cases. In each sub-case, we prove that ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}} \le {\mathfrak {M}} \wedge \mathfrak {M^{*}} \le {\mathfrak {M'}} \wedge v_\mathfrak {M^{*}}$(A) $= v_\mathfrak {M^{*}}$(C $\vee$ D) $= v_\mathfrak {M^{*}}$(B) $= n$).

• Sub-case 2a$v_{\mathfrak {M}}$(C) $= n$ but $v_{\mathfrak {M}}$(D) $= 0$.

  In this case, $v_{\mathfrak {M}}$(C) $= n$ and $v_{\mathfrak {M'}}$(B) $= n$. So, by inductive hypothesis, ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}} \le {\mathfrak {M}} \wedge \mathfrak {M^{*}} \le {\mathfrak {M'}} \wedge v_\mathfrak {M^{*}}$(C) $= v_\mathfrak {M^{*}}$(B) $= n$). Since $\mathfrak {M^{*}} \le {\mathfrak {M}}$, it follows that $v_\mathfrak {M^{*}}$(D) $\le v_{\mathfrak {M}}$(D) by Proposition 1. Since $v_{\mathfrak {M}}$(D) $= 0$ and $v_\mathfrak {M^{*}}$(D) $\le v_{\mathfrak {M}}$(D), it further follows that $v_\mathfrak {M^{*}}$(D) $= 0$ or $v_\mathfrak {M^{*}}$(D) $= n$. Either way, ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($\mathfrak {M^{*}} \le {\mathfrak {M}} \wedge \mathfrak {M^{*}} \le {\mathfrak {M'}} \wedge v_\mathfrak {M^{*}}$(A) $= v_\mathfrak {M^{*}}$(C $\vee$ D) $= v_\mathfrak {M^{*}}$(B) $= n$).

• Sub-case 2b$v_{\mathfrak {M}}$(C) $= 0$ but $v_{\mathfrak {M}}$(D) $= n$.

  The proof of sub-case 2b is similar to that of sub-case 2a.

• Sub-case 2c$v_{\mathfrak {M}}$(C) $= n$ but $v_{\mathfrak {M}}$(D) $= n$.

  The proof of sub-case 2c is similar to that of sub-case 2a. $\square$

Lemma 3

(($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $\ne 1$) $\wedge$ ($\exists {\mathfrak {M'}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $\ne 0$)) $\rightarrow$ ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($v_\mathfrak {M^{*}}$(A) $= n$), for any sentence A.

Proof

We prove Lemma 3 by induction on the number of connectives in A. Again for simplicity, we assume that ‘$\wedge$’ is defined in terms of ‘$\lnot$’ and ‘$\vee$’ and we omit the quantificational case.

• Base case This is automatically true by Assumption (B).

• Inductive step Assume that Lemma 3 is true of sentences B and C whose numbers of connectives are less than that in A. Two cases:

• Cases 1 A is ‘$\lnot$B’

  Assume that ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= v_{\mathfrak {M}}$($\lnot$B) $\ne 1$) and ($\exists {\mathfrak {M'}} \in {\mathbb {S}}$)($v_{\mathfrak {M'}}$(A) $= v_{\mathfrak {M'}}$($\lnot$B) $\ne 0$). Then ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 0$) and ($\exists {\mathfrak {M'}} \in {\mathbb {S}}$)($v_{\mathfrak {M'}}$(B) $\ne 1$). So ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($v_\mathfrak {M^{*}}$(B) $= n$) by inductive hypothesis. So ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($v_\mathfrak {M^{*}}$(A) $= v_\mathfrak {M^{*}}$($\lnot$B) $= v_\mathfrak {M^{*}}$(B) $= n$).

• Case 2 A is ‘B $\vee$ C’

  Assume that ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= v_{\mathfrak {M}}$(B $\vee$ C) $\ne 1$) and ($\exists {\mathfrak {M'}} \in {\mathbb {S}}$)($v_{\mathfrak {M'}}$(A) $= v_{\mathfrak {M'}}$(B $\vee$ C) $\ne 0$). By the fact that ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= v_{\mathfrak {M}}$(B $\vee$ C) $\ne 1$), it follows that ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 1$) and ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(C) $\ne 1$). And, by the fact that ($\exists {\mathfrak {M'}} \in {\mathbb {S}}$)($v_{\mathfrak {M'}}$(A) $= v_{\mathfrak {M'}}$(B $\vee$ C) $\ne 0$), it follows that either ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 0$) or ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(C) $\ne 0$). We prove by cases in what follows that either case leads to the conclusion that ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($v_\mathfrak {M^{*}}$(A) $= v_\mathfrak {M^{*}}$(B $\vee$ C) $= n$).

• Case 2a ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 0$)

  In this case, ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 1$) and ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 0$). So, by the inductive hypothesis, ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($v_\mathfrak {M^{*}}$(B) $= n$). Now, ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(C) $\ne 1$). So, either ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(C) $= 0$) or ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(C) $= n$), otherwise it will contradict the inductive hypothesis that the lemma holds for C. If ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(C) $= 0$), then the model $\mathfrak {M^{*}} \in {\mathbb {S}}$ such that $v_\mathfrak {M^{*}}$(B) $= n$ will also be a model in which $v_\mathfrak {M^{*}}$(A) $= v_\mathfrak {M^{*}}$(B $\vee$ C) $= n$. And if ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(C) $= n$), then, since ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($v_\mathfrak {M^{*}}$(B) $= n$), there will be a model ${\mathfrak {M'}}$ such that $v_{\mathfrak {M'}}$(B) $= v_{\mathfrak {M'}}$(C) $= n$ by Lemma 2 and therefore $v_{\mathfrak {M'}}$(A) $= v_{\mathfrak {M'}}$(B $\vee$ C) $= n$. So, if ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 0$), then ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($v_\mathfrak {M^{*}}$(A) $= v_\mathfrak {M^{*}}$(B $\vee$ C) $= n$).

• Case 2b ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(C) $\ne 0$)

  The proof of Case 2b is similar to that of Case 2a.

Now we prove the main theorem: if a disjunction is true simpliciter, then at least one of its disjuncts is true simpliciter, i.e.:

Theorem 1

If ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A $\vee$ B) $= 1$), then ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= 1$) or ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $= 1$).

Proof

Assuming that ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A $\vee$ B) $= 1$), we prove the consequent of the theorem by reductio. Suppose that it is neither the case that ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= 1$) nor the case that ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $= 1$). So, ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $\ne 1$) and ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 1$). Four possibilities:

• Possibility 1 ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= 0$) and ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $= 0$). In this case, ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A $\vee$ B) = 0), which contradicts our initial assumption.

• Possibility 2 ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= 0$) but not $(\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $= 0$). In this case, ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 0$). Since ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 1$) and ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 0$), it follows by Lemma 3 that ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $= n$). So, ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $= n\,\wedge \,v_{\mathfrak {M}}$(A) $= 0$). Therefore, ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A $\vee$ B) $= n$), which contradicts our initial assumption.

• Possibility 3 ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $= 0$) but not $(\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= 0$). In this case, ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $\ne 0$). Since ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $\ne 1$) and ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $\ne 0$), it follows by Lemma 3 that ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= n$). So, ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= n$ and $v_{\mathfrak {M}}$(B) $= 0$). Therefore, ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A $\vee$ B) $= n$), which contradicts our initial assumption.

• Possibility 4 Neither ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= 0$) nor ($\forall {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $= 0$). So ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $\ne 0$) and ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 0$). It follows from the facts that ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $\ne 1$) and ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $\ne 0$), by Lemma 3, that ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= n$). Similarly, it follows from the facts that ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 1$) and ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(B) $\ne 0$), again by Lemma 3, that ($\exists {\mathfrak {M'}} \in {\mathbb {S}}$)($v_{\mathfrak {M'}}$(B) $= n$). So, ($\exists {\mathfrak {M}} \in {\mathbb {S}}$)($v_{\mathfrak {M}}$(A) $= n$) and ($\exists {\mathfrak {M'}} \in {\mathbb {S}}$)($v_{\mathfrak {M'}}$(B) $= n$). By Lemma 2, it follows that ($\exists \mathfrak {M^{*}} \in {\mathbb {S}}$)($v_\mathfrak {M^{*}}$(A) $= v_\mathfrak {M^{*}}$(B) $= n = v_\mathfrak {M^{*}}$(A $\vee$ B)), which contradicts our initial assumption.