Proof: All bases are triangular Lemma: Some row has exactly one basic variable Assume each row has at least two basic variables Let k be the number of bas variables Then there are at least 2n basic variables in the supplies section, and at least 2m basic variables in the demands section thus k >2n. k>2m, so k >m+n. But since there are only m+n- independent equations, there is a contradiction Proof: All bases are triangular Lemma: Exclusion of any redundant equation from the original system leaves an equation in exactly one basic variable Drop some equation as redundant, say, the last demand equation Let k' be the new number of basic variables. Again make the contrary assumption to get k>2(n-1) k>2m. There was at least one basic entry in the equation excluded, so k >k+I. Then k>m+n-1/2, which again contradicts k <m+n-1 Then, successively delete the row having a single basic variable and repeat the argument for the reduced array, establishing the theoremProof: All bases are triangular • variable variables. Let k be the number of basic variables. Then there are at least 2n basic variables in the supplies section, and at least 2m basic variables in the demands section, thus k >2n , k >2m , so k >m+n. But since there are only m+n-1 independent equations, there is a contradiction. Proof: All bases are triangular • from the original system leaves an equation in exactly one basic variable. – equation. Let k’ be the new number of basic variables. Again make the contrary assumption to get k’ >2(n-1) , k >2m. There was at least one basic entry in the equation excluded, so k >k’+1. Then k >m+n-1/2, which again contradicts k <m+n-1. • basic variable and repeat the argument for the reduced array, establishing the theorem. Lemma: Some row has exactly one basic – Assume each row has at least two basic Lemma: Exclusion of any redundant equation Drop some equation as redundant, say, the last demand Then, successively delete the row having a single