Let L be the set of all normal and convex functions from [0,1] to [0,1]. This paper proves that t-norm in the sense of
Walker-and-Walker is strictly stronger that trβ-norm on L, which
is strictly stronger than t-norm on L. Furthermore, let
β and β be special convolution operations defined by
(fβg)(x)=sup{f(y)βg(z):yβ³z=x},(fβg)(x)=sup{f(y)βg(z):yΒ β½Β z=x}, for f,gβMap([0,1],[0,1]), where β³
and β½ are respectively a t-norm and a t-conorm on [0,1] (not necessarily continuous), and β is a binary operation on [0,1]. Then, it is proved that if the binary operation β is a
trβ-norm (resp., β is a trβ-conorm), then β³
is a continuous t-norm (resp., β½ is a continuous
t-conorm) on [0,1], and β is a t-norm on [0,1].Comment: arXiv admin note: text overlap with arXiv:1908.10532,
arXiv:1907.1239