axiom f x == a := 3 x + a Type: Void
axiom f 3 axiom Compiling function f with type PositiveInteger -> PositiveInteger
Type: PositiveInteger?
axiom g := x +-> (x+2 ; x+a)
Type: AnonymousFunction?
axiom g 3
Type: Polynomial(Integer)
are right functions.
But I can't use the function h. I get an error - axiom h := x +-> (a := 3 ; x+a)
Type: AnonymousFunction?
axiom h 3
Type: PositiveInteger?
... --gdr, Thu, 24 Jan 2008 21:23:23 -0800 reply fixed in OpenAxiom
Status: open => fixed somewhere
Status: fixed somewhere => closed
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last edited 9 years ago by test1 |
![\label{eq2}x \mapsto{
\begin{array}{l}
{\left(x + 2 \right)}
\
{\left(x + a \right)}](images/1972558510237511019-16.0px.png "\label{eq2}x \mapsto{
\begin{array}{l}
{\left(x + 2 \right)}
\
{\left(x + a \right)}")
![\label{eq4}x \mapsto{
\begin{array}{l}
{\left(a : = 3 \right)}
\
{\left(x + a \right)}](images/7071648393611172993-16.0px.png "\label{eq4}x \mapsto{
\begin{array}{l}
{\left(a : = 3 \right)}
\
{\left(x + a \right)}")
