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Proof step 13 from proof min-le-diff-third (previous step)

n2 ≤ n2 - (n0 < (n1 < n2 ⊃ n1;n2) ⊃ n0;(n1 < n2 ⊃ n1;n2)) ⊦ n0 = 0, n1 = 0, n2 = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 14:
n0 < (n1 < n2 ⊃ n1;n2), n2 ≤ n2 - n0 ⊦ n0 = 0, n1 = 0, n2 = 0
Proof step 15:
n2 ≤ n2 - (n1 < n2 ⊃ n1;n2) ⊦ n0 = 0, n1 = 0, n2 = 0, n0 < (n1 < n2 ⊃ n1;n2)

Rule argument

pos: left 1

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