let p be polyhedron; :: thesis: for k being Integer st k = - 1 holds
dim (k -bounding-chain-space p) = 1

let k be Integer; :: thesis: ( k = - 1 implies dim (k -bounding-chain-space p) = 1 )
set T = 0 -boundary p;
set V = k -bounding-chain-space p;
assume A1: k = - 1 ; :: thesis: dim (k -bounding-chain-space p) = 1
card ([#] (k -bounding-chain-space p)) = 2
proof end;
hence dim (k -bounding-chain-space p) = 1 by RANKNULL:6; :: thesis: verum