+ // When trying to find a pivot, we should try a bit harder than expand().
+ // Searching the first non-zero element in-place here instead of calling
+ // pivot() allows us to do no more substitutions and back-substitutions
+ // than are actually necessary.
+ int indx = r0;
+ while ((indx<m) &&
+ (tmp_n[indx*n+c0].subs(srl, subs_options::no_pattern).expand().is_zero()))
+ ++indx;
+ if (indx==m) {
+ // all elements in column c0 below row r0 vanish