This code does it:

abCaseSet := {[`0<a<1`, `0<b<1`], [`0<a<1`, `b=-1`], [`0<a<1`, `b=0`], [`0<a<1`, `b=1`], [`0<a<1`, `1<b<&infin;`], [`0<a<1`, -`1<b<0`], [`0<a<1`, -`&infin;<b<-1`], [`a=-1`, `0<b<1`], [`a=-1`, `b=-1`], [`a=-1`, `b=0`], [`a=-1`, `b=1`], [`a=-1`, `1<b<&infin;`], [`a=-1`, -`1<b<0`], [`a=-1`, -`&infin;<b<-1`], [`a=0`, `0<b<1`], [`a=0`, `b=-1`], [`a=0`, `b=0`], [`a=0`, `b=1`], [`a=0`, `1<b<&infin;`], [`a=0`, -`1<b<0`], [`a=0`, -`&infin;<b<-1`], [`a=1`, `0<b<1`], [`a=1`, `b=-1`], [`a=1`, `b=0`], [`a=1`, `b=1`], [`a=1`, `1<b<&infin;`], [`a=1`, -`1<b<0`], [`a=1`, -`&infin;<b<-1`], [`1<a<&infin;`, `0<b<1`], [`1<a<&infin;`, `b=-1`], [`1<a<&infin;`, `b=0`], [`1<a<&infin;`, `b=1`], [`1<a<&infin;`, `1<b<&infin;`], [`1<a<&infin;`, -`1<b<0`], [`1<a<&infin;`, -`&infin;<b<-1`], [-`1<a<0`, `0<b<1`], [-`1<a<0`, `b=-1`], [-`1<a<0`, `b=0`], [-`1<a<0`, `b=1`], [-`1<a<0`, `1<b<&infin;`], [-`1<a<0`, -`1<b<0`], [-`1<a<0`, -`&infin;<b<-1`], [-`&infin;<a<-1`, `0<b<1`], [-`&infin;<a<-1`, `b=-1`], [-`&infin;<a<-1`, `b=0`], [-`&infin;<a<-1`, `b=1`], [-`&infin;<a<-1`, `1<b<&infin;`], [-`&infin;<a<-1`, -`1<b<0`], [-`&infin;<a<-1`, -`&infin;<b<-1`]};

map([proc(L)
local i;
    seq(
    (proc(x, f, y)
      if member(f, {`<`, `<=`, `=`, `>`, `>=`}) then f(x, y) end if;
    end proc)(op(i - 1 .. i + 1, L)), i = 2 .. nops(L) - 1)
  end proc], parse(
  (proc(X)
local L, S;
    S := convert(X, string);
    for L in [["`", ""], ["&infin;", "infinity"], ["<=", ", `<=`, "],
      ["<", ", `<`, "], [">=", ", `>=`, "], [">", ", `>`, "], ["=", ", `=`, "]
      ] do S := StringTools:-SubstituteAll(S, op(L))
    end do
  end proc)(abCaseSet)));

@Luca3544 Are you trying to make your procedure to do a print out of a search? If so then attach "NULL;" as last statement to the procedure code.

M := proc(a)
local Grundstof, c;
description "Side 302, 303";
  c := convert(a, string);
  Grundstof := [(1.0079*g)/(mol) + "H", (4.0026*g)/(mol) + "He",
  (6.941*g)/(mol) + "Li", (9.0122*g)/(mol) + "Be", (g)/(mol),
  (10.811*g)/(mol) + "B", (12.0107*g)/(mol) + "C", (14.0067*g)/(mol) + "N",
  (15.9994*g)/(mol) + "O", (18.9984*g)/(mol) + "F", (22.990*g)/(mol) + "Na",
  (24.305*g)/(mol) + "Mg"];
  # (proc(E, n, s) if has(E, s) then print(E); n fi end)~(Grundstof, [$1..nops(Grundstof)], c);
  # or
  map(proc(n, c) if has(args) then print(n) fi end, Grundstof, c);
  NULL;
end proc;

That should be "assuming 0 < k[2];"

I do not believe you can define an assumption like "for all i k[i] is positive".

f1 := unapply(diff(g1, x), x);

or

gf := unapply(g1, x);
f1 := D(gf);

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This should be:

geometry:-RegularPolygon(gon,5,geometry:-point(o,1,1),2);
#or
use geometry in RegularPolygon(gon,5,point(o,1,1),2) end use;

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Deleted.

If this does not work please let me know:

 F1 := (X, Y) -> [X, Y+abs(X*K+sin(ap))/abs((X*K+sin(ap))^2-1)^(1/2)];

plottools[transform](F1)(p1);

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First convert to piecewise. D(abs)(x) is defined as abs(1, x) and is undefined at x=0. Avoid 0 of the abs function like this:
 

collect(diff(I2, c), abs, normal);
evalindets(%, 'abs(1, anything)', x -> normal(convert(x, piecewise, c))) assuming (c<>-1, c<>-0.9, c<>.9000000000, c<>1);

#or

diff(convert(I2, piecewise, c), c):
map(collect, normal(%), ln, normal@simplify) assuming (c<>-1, c<>-0.9, c<>.9000000000,c<>1);

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I am assuming that the answer I got from Maple is correct.

restart;
with(Student:-MultivariateCalculus);
A := [0, 0, 0];
B := [c, 0, 0];
S := [x, 0, z];
solve([Distance(S, A) = a, Distance(S, B) = b], [x, z], explicit = false) assuming (a > 0, b > 0, c > 0);

Perhaps the answer expected is meant to include the domain of real solutions (?). Then:

restart;
with(Student:-MultivariateCalculus);
A := [0, 0, 0];
B := [c, 0, 0];
S := [x, 0, z];
Q := a > 0, b > 0, c > 0;
solve([Distance(S, A) = a, Distance(S, B) = b], [x, z], explicit = true);
ANS := (R -> [R, map(solve @ `union`, [solve]([op(evalc(subs(R, [Im(x), Im(z)])) assuming Q), Q]), {Q}) ]) ~ (%);
%[1];

The result given is:
[x = (1/2)*(a^2-b^2+c^2)/c, z = (1/2)*(-a^4+2*a^2*b^2+2*a^2*c^2-b^4+2*b^2*c^2-c^4)^(1/2)/c]
[{a = -b+c, 0 < b, 0 < c, b < c}, {0 < b, 0 < c, a < b+c, b < c, -b+c < a}, {a = b+c, 0 < b, 0 < c, b < c}, {b = c, 0 < a, 0 < c, a < 2*c}, {a = 2*c, b = c, 0 < c}, {a = b-c, 0 < c, c < b}, {0 < c, a < b+c, c < b, b-c < a}, {a = b+c, 0 < c, c < b}]
[x = (1/2)*(a^2-b^2+c^2)/c, z = -(1/2)*(-a^4+2*a^2*b^2+2*a^2*c^2-b^4+2*b^2*c^2-c^4)^(1/2)/c]
[{a = -b+c, 0 < b, 0 < c, b < c}, {0 < b, 0 < c, a < b+c, b < c, -b+c < a}, {a = b+c, 0 < b, 0 < c, b < c}, {b = c, 0 < a, 0 < c, a < 2*c}, {a = 2*c, b = c, 0 < c}, {a = b-c, 0 < c, c < b}, {0 < c, a < b+c, c < b, b-c < a}, {a = b+c, 0 < c, c < b}

From the above we have abs(b-c) < a < b+c.

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Haven't you answered this one in previous posts already? Try something like this:

E := {x, sin(x), ln(y), ln(x+y), exp(z)};
P := randpoly(E, degree=4,terms=8);
F := P -> frontend(proc(E, x) local c; [coeffs(collect(E, x, distributed), x, 'c')], [c] end, [P, indets(P, {'ln(anything), exp(anything)'})], [{Non}(function), {}]);
F(P);
F(expr);

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I am guessing, but this might be the only way:

null_record:=Record(name, age);
foo:=proc() #some proc that uses same Record layout
global myrecord;
   local S:=Record[null_record]();
   S:-age:=20;
   return S;
end proc:

You may want to look at packed option.

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Never mess with builtin functions like D. Use d instead.

restart: with(plots):
alias(c[0]=c0, c[1]=c1, c[2]=c2);
PDE:=diff(C(x,t),t)=d*diff(C(x,t),x,x);
IBC:={C(x,0)=cx0, C(0,t)=ct0, D[1](C)(10,t)=0};
ct0:=1;
cx0:=0;
d:=1;
pds:=pdsolve(PDE,IBC,numeric);
L1:=[0.01, 0.1, 1, 5, 10];
L2:=[red, green, yellow,  blue, magenta, black];
for i from 1 to 5 do
 pn[i] := pds:-plot(t=L1[i], color=L2[i]):
end do:
display({seq(pn[i], i=1..5)}, title=`Numerical solution at t=0.01, 0.1, 1, 5, 10`);

So that is supposed to be the RLC circuit?
If you need then this is from google: https://en.wikipedia.org/wiki/RLC_circuit

L*(diff(x(t), t, t)) + R*(diff(x(t), t)) + (x(t))/(C) = 0;
dsolve(%);

I presume all the other questions You posted are solve the same.

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subsop(1=1, expr); #for multiplication
subsop(1=0, expr); #for addition
subsop(1=NULL, expr); #for both (?)

That should be M[i,j] := 1;  instead of M[i][j] := 1;