theorem Th6:
  a|^m + b|^m <= c|^m implies a <= c
  proof
    assume
    A0: a|^m + b|^m <= c|^m;
    m=0 implies a|^m + b|^m > c|^m by Lm0; then
    A2e: 0+1 <= m by A0,NAT_1:13;
    a|^m + b|^m <= c|^m + b|^m by A0,NAT_1:12;
    then a|^m + b|^m -b|^m <= c|^m + b|^m- b|^m by XREAL_1:9;
    hence thesis by A2e,PREPOWER:10;
  end;
