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Proof step 7 from proof le-03 (previous step)

m +1 ≠ 0 → (m +1 * n ≤ m +1 * n0 ↔ n ≤ n0) ⊦ m +1 * n ≤ m +1 * n0 ↔ n ≤ n0

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic