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Proof step 2 from proof modls (previous step)

n0 < m0 ⊦ m0 * m + n0 < m0 * n ↔ m < n

User applied rule case distinction

And obtained the following new goals

Proof step 3:
m0 * m + n0 < m0 * n, n0 < m0 ⊦ m < n
Proof step 5:
m < n, n0 < m0 ⊦ m0 * m + n0 < m0 * n

Rule argument

pos: right 1

Additional information