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Proof step 2 from proof le (previous step)

⊦ m0 * m ≤ m0 * n ↔ m ≤ n, m0 = 0

User applied rule case distinction

And obtained the following new goals

Proof step 3:
m0 * m ≤ m0 * n ⊦ m ≤ n, m0 = 0
Proof step 6:
m ≤ n ⊦ m0 * m ≤ m0 * n, m0 = 0

Rule argument

pos: right 1

Additional information