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Proof step 2 from proof le-succ (previous step)

⊦ m * n ≤ m + 1 ↔ m = 0 ∨ m = 1 ∧ n ≤ 2 ∨ 2 ≤ m ∧ n < 2

User applied rule case distinction

And obtained the following new goals

Proof step 3:
m * n ≤ m + 1 ⊦ m = 0 ∨ m = 1 ∧ n ≤ 2 ∨ 2 ≤ m ∧ n < 2
Proof step 6:
m = 0 ∨ m = 1 ∧ n ≤ 2 ∨ 2 ≤ m ∧ n < 2 ⊦ m * n ≤ m + 1

Rule argument

pos: right 1

Additional information