Answer to Question #234706 in Math for smi

Question

For any t-conorm ∇ show that ∇₅≤∇≤∇₀.

Answer 1

Proof

We have ∇₀ and ∇_s t-conorms; Δ₀ and Δ_s t-norms

For each t-norm Δ₀ and $x \epsilon [0,1]$ ; the boundary conditions are

$Δ_o(x,1)= ∇_0(1,x)=x\\ Similarly\\ Δ_5(x,0)= ∇_0(0,x)=x\\$

Monotonicity in both components;

$Δ_5(x_1 , y_1) ≤Δ_0(x_2,y_2)=x\\ where x_1 ≤ x_2, y_1 ≤ y_2$

From Lukasiewicz t norm T_l

we get Δ_l (a,b) = max {a+b-1,0}

for all $a,b\space \epsilon [0,1]\\ Also, ∇_s ≤ ∇_0$ , so from here we get ∇₅≤∇₀.

Hence ∇₅≤∇≤∇₀