Fuzzy Arithmetic - Fuzzy Logic: An Introductory Course for Engineering Students - page 143

Namely, take

and

(μ n ↕ μ m )( + ) = ≤ + − (ʱ + ʲ) (μ n ↕ μ m )( ) = ;

Obviously n m 1; if t m n , t 0

if t ≥ m + n + (ʱ + ʲ) , (μ n ↕ μ m )( t ) = 0. Hence, the problem is reduced to com-

pute the values of μ n ↕ μ m in the two intervals ( m + n − (ʱ + ʲ), m + n ) , and

( m + n , m + n + (ʱ + ʲ)) . It can be done as follows:

1. If x ∈ ( n − ʱ, n ) , and y ∈ ( m − ʲ, m ) , t = x + y ∈ ( n + m − (ʱ + ʲ), n + m ) , and,

x − n y + m

since μ n ( x ) = 1 + , μ m ( y ) = 1 + , it follows (with μ n ( x ) = μ m ( y ) = z ):

ʱ ʲ

x = ʱ z + n − ʱ , y = ʲ z + m − ʲ , and

t = x + y = (ʱ + ʲ) z + n + m − ʱ − ʲ,

giving

t − ( n + ) + (ʱ + ʲ)

ʱ + ʲ

m

z = ,

that is

− ( + ) + (ʱ + ʲ)

ʱ + ʲ

t n m

(μ n ↕ μ m )( ) = , ∈ ( + − (ʱ + ʲ), + ).

t if t m n m n

2. If x ∈ ( n , n + ʱ) , and y ∈ ( m , m + ʲ) , t ∈ ( n + m , n + m + (ʱ + ʲ)) , and like

t − ( n + ) + (ʱ + ʲ)

ʱ + ʲ

m

in (1), it is z = , that is

t − ( n + ) − (ʱ + ʲ)

ʱ + ʲ

m

(μ n ↕ μ m )( t ) = , if t ∈ ( n + m , m + n − (ʱ + ʲ)).

3. Graphically, with μ n ↕ μ m = μ n + m