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Proof step 5 from proof alignup-add-le (previous step)

(n0 % m = 0 ⊃ n0;m * n0 / m + m) + n1 ≤ n0 ⊦ m = 0, n1 = 0 ∧ n0 % m = 0

User applied rule case distinction

And obtained the following new goals

Proof step 6:
(m * n0 / m + m) + n1 ≤ n0 ⊦ n0 % m = 0, m = 0, n1 = 0 ∧ n0 % m = 0

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic