1次結合の等式は 0ˉ=x(aˉ+bˉ)+y(bˉ+cˉ)+z(cˉ+aˉ)\bar{0} = x(\bar{a} + \bar{b}) + y(\bar{b} + \bar{c}) + z(\bar{c} + \bar{a})0ˉ=x(aˉ+bˉ)+y(bˉ+cˉ)+z(cˉ+aˉ) =(x+z)aˉ+(x+y)bˉ+(y+z)cˉ= (x + z)\bar{a} + (x + y)\bar{b} + (y + z)\bar{c}=(x+z)aˉ+(x+y)bˉ+(y+z)cˉ aˉ,bˉ,cˉ\bar{a}, \bar{b}, \bar{c}aˉ,bˉ,cˉ は1次独立なので x+z=x+y=y+z=0x + z = x + y = y + z = 0x+z=x+y=y+z=0 ∴x=y=z=0\therefore x = y = z = 0∴x=y=z=0 よって (aˉ+bˉ),(bˉ+cˉ),(cˉ+aˉ)(\bar{a} + \bar{b}), (\bar{b} + \bar{c}), (\bar{c} + \bar{a})(aˉ+bˉ),(bˉ+cˉ),(cˉ+aˉ) は1次独立となる