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Solutions

For fuzzy-logic-consequence we check all valuations to see if the smallest premise value will always be lower than the conclusion value. The idea is that this shows us that whenever the premises get ‘closer’ to $1$, the conclusion does too, so there is a truth-strengthening connection between them.

(a) This is invalid. Consider a valuation where $R$ is fuzzy or vague, $v(R)=0.5$, but the conclusion is definitely false $v(H)=0$. Then we have $v(\neg R)=0.5$ and $v(R\land\neg R)=0.5$ which is greater than the conclusion value.

(b) This is valid. We want to show that we always get: $v(H)\leq v(H\lor(R\land H))$.

• Break down the calculation $v(H\lor(R\land H))=\mathrm{max}(v(H),\mathrm{min}(v(R),v(H)))$.
• Consider two ways the atomic parts might be interpreted: $v(H)\leq v(R)$ or $v(H)>v(R)$. Every valuation has to be one of these types. We will show that the result holds on either type.
• If we are looking at the first type of valuation, we have $\mathrm{min}(v(R),v(H))=v(H)$, but that makes $\mathrm{max}(v(H),\mathrm{min}(v(R),v(H)))=\mathrm{max}(v(H),v(H))=v(H)$. So we have $\mathrm{max}(v(H),\mathrm{min}(v(R),v(H)))\leq v(H)$ as desired.
• If we are looking at the second type of valuation, we have $\mathrm{min}(v(R),v(H))=v(R)$, but that makes $\mathrm{max}(v(H),\mathrm{min}(v(R),v(H)))=\mathrm{max}(v(H),v(R))=v(H)$. Again, we have $\mathrm{max}(v(H),\mathrm{min}(v(R),v(H)))\leq v(H)$ as desired.

(c) This is invalid. Consider a valuation where $U$ is fuzzy or vague, $v(U)=0.5$, but the other statements are definitely false $v(H)=0$ and $v(D)=0$. Then we have $v(H\lor\neg U)=\mathrm{max}(v(H),1-v(U))=0.5$. We also have $v(U\lor D)=\mathrm{max}(v(U),v(D))=0.5$. Take the minimum $\mathrm{min}(\mathrm{max}(v(H),v(\neg U)),\mathrm{max}(v(U),v(D)))=0.5$. But this value is greater than the conclusion value $v(H\lor D)=\mathrm{max}(v(H),v(D))=0$, so we have a counter-model.

(d) This is valid. We want to show that we always get: $v(H\land\neg H)\leq v(H)$. It’s immediate from $v(H\land\neg H)=\mathrm{min}(v(H),1-v(H))$. This guarantees that the premises cannot exceed the value of H itself.

• To illustrate, suppose that $v(H)=0$ then we get $v(H\land\neg H)= 0$.
• Suppose that $v(H)=0.1$ then we get $v(H\land\neg H)= 0.1$.
• Suppose that $v(H)=0.5$ then we get $v(H\land\neg H)= 0.5$.
• Suppose that $v(H)=0.9$ then we get $v(H\land\neg H)= 0.1$.
• Suppose that $v(H)=1$ then we get $v(H\land\neg H)= 0$.

Solutions

(a) Here is the best way to write the rules in our limited formal language. Remember, we want to interpret these like fuzzy-logic-consequence: whenever the antecedent side gets ‘closer’ to $1$, we want the conclusion to do the same.

IF HOT$\lor$HUMID, THEN AC

IF $\neg$HOT$\lor$HUMID, THEN HEATER

IF AC, THEN $\neg$HEATER

IF HEATER, THEN $\neg$AC

(b) One example of a problem for these rules: situations with low heat and high humidity. To be precise, consider a valuation where $v(\mathsf{HOT})=0.2$ and $v(\mathsf{HUMID})=0.8$. This makes $v(\neg\mathsf{HOT})=1-v(\mathsf{HOT})=0.8$.

We calculate $v(\mathsf{HOT}\lor\mathsf{HUMID})=0.8$.

So the first rule requires $0.8\leq v(\mathsf{AC})$.

We also calculate $v(\neg\mathsf{HOT}\lor\mathsf{HUMID})=0.8$.

So the second rule requires $0.8\leq v(\mathsf{HEATER})$.

This is okay so far, but the final rules create a problem.

For instance, the last rule requires $0.8\leq 1-v(\mathsf{AC})$.

There is no value that gives $0.8\leq v(\mathsf{AC})$ and $0.8\leq 1-v(\mathsf{AC})$.

So these rules do not ‘work’ as intended in all possible fuzzy valuations.

Solution

(a) Strong Conjunction.

The $f_\otimes$ function calculates its output in two different ways, depending on the input values. We get a default value $f_\otimes(x,y)=0$ when $x+y\leq 1$, but we get $f_\otimes(x,y)=x+y-1$ when $1<x+y$. For these exercises we prove both cases separately.

(1) We prove the goal in two cases using $x+y=y+x$.

1. If $f_\otimes(x,y)=0$ then $x+y\leq 1$. This implies that $y+x\leq 1$ and so $f_\otimes(y,x)=0$.

2. If $f_\otimes(x,y)\neq 0$ then $f_\otimes(x,y)=x+y-1=y+x-1=f_\otimes(y,x)$.

(2) We prove the goal in two cases.

1. If $f_\otimes(x,y)=0$ then obviously $f_\otimes(x,y)\leq f_\otimes(x’,y’)$.

2. If $f_\otimes(x,y)\neq 0$ then we reason:

• $x\leq x’$ and $y\leq y’$
• so we have $x+y\leq x’+y’$
• so we have $x+y-1\leq x’+y’-1$
• therefore $f_\otimes(x,y)\leq f_{\oplus}(x’,y’)$

(3) We prove the goal in two cases using $x+(y+z)=(x+y)+z$.

1. If $f_\otimes(x,f_\otimes(y,z))=0$ then SOMETHING

2. If $f_\otimes(x,y)\neq 0$ then we reason: SOMETHING

(4) We prove the goal in two cases.

1. If $f_\otimes(x,1)=0$ this means that $x+1\leq 1$ and so $0=x$.

2. If $f_\otimes(x,1)\neq 0$ then $f_\otimes(x,1)= x+1-1 = x$.

(b) Fuzzy Conditional.

The $f_\to$ function calculates its output in two different ways, depending on the input values. We default to the antecedent value $f_\to(x,y)=x$ when $2x-y\leq 1$, but we get $f_\to(x,y)=1-x+y$ when $1<2x-y$.

For fuzzy-logic-consequence we check all valuations to see if the smallest premise value will always be lower than the conclusion value. So we want to show that we always get: $v(R\otimes (R\to W))\leq v(W)$.

Break down the calculation $v(R\otimes (R\to W))=f_\otimes(v(R),v(R\to W))=f_\otimes(v(R),f_\to(v(R),v(W)))$.

We can think about how $f_\otimes(v(R),f_\to(v(R),v(W)))$ is calculated in two ways.

1. The first possibility is quick: If $f_\otimes(v(R),f_\to(v(R),v(W)))=0$ then obviously $v(R\otimes (R\to W))\leq v(W)$.
2. The second possibility requires more thinking: suppose that $f_\otimes(v(R),f_\to(v(R),v(W)))\neq 0$.

• That means $f_\otimes(v(R),f_\to(v(R),v(W)))=v(R)+f_\to(v(R),v(W))-1$.
• We now think about how $f_\to(v(R),v(W))$ is calculated in two ways.
• First, if $2v(R)-v(W)\leq 1$ that gives us two pieces of information. We have $f_\to(v(R),v(W))=v(R)$ which plugs into our original calculation to give us $f_\otimes(v(R),f_\to(v(R),v(W)))=2v(R)-1$ and we also have $2v(R)-1\leq v(W)$. It follows that $f_\otimes(v(R),f_\to(v(R),v(W)))\leq v(W)$.
• Second, if $1<2v(R)-v(W)$, then $f_\to(v(R),v(W))=1-v(R)+v(W)$ and $f_\otimes(v(R),f_\to(v(R),v(W)))=v(R)+(1-v(R)+v(W))-1=v(W)$. It follows that $f_\otimes(v(R),f_\to(v(R),v(W)))\leq v(W)$.