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Proof step 4 from proof le-same (previous step)

m * n + n0 ≤ m ⊦ n = 1 ∧ n0 = 0, m = 0 ∧ n0 = 0, n = 0 ∧ n0 ≤ m

User applied rule case distinction

And obtained the following new goals

Proof step 5:
m * n + n0 ≤ m ⊦ n = 0, m = 0 ∧ n0 = 0, n = 1 ∧ n0 = 0

Rule argument

pos: right 3

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic