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Let ${\ell }_N(x)=3.44-x$ and ${\ell }_A(x)=-7.59-3x$. The common-potential test for a three-mineral activity diagram gives relative potentials, in one positive energy unit,The stable phase minimizes these potentials. Consequently the stable nepheline–corundum branch has $x\le -5.515$, the stable albite–corundum branch has $x\ge -5.515$, and the stable nepheline–albite branch has $x=-5.515$, $y\ge 8.955$. Draw them solid.

The other halves of each line are metastable: a third mineral has lower potential there. In particular, below the triple point the vertical equality describes two phases both less stable than corundum. Extend those halves as dashed lines. This thermodynamic comparison prevents the incorrect use of all three complete pairwise lines as stable field boundaries.

A stability boundary in a mineral activity diagram must also be stable against the third mineral. Use the common-potential test for a three-mineral activity diagram. For the two corundum-producing directions,Negative values favour corundum. For the albite-to-nepheline conversion, ${\Delta }_{r}G/(RT\ln 10)=2x+11.03$, so low silica activity favours nepheline and high silica activity favours albite. These comparisons give:

All three coexist at the intersection. Pairwise mass-action equations alone do not distinguish the stable segments from their metastable extensions.