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Proof step 7 from proof alignup-add-bounded1 (previous step)

n ≤ m ⊦ n % m = 0, m * n / m + m = (n = 0 ⊃ 0;m), m = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
n ≤ m ⊦ n % m = 0, m * n / m + m = m, m = 0, n = 0

Rule argument

pos: right 2

Additional information

Used simplifier rules: n ≠ 0 ⊦ 0 / n = 0; (s)% ⊦ 0 * n = 0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≠ 0 → 0 % m = 0; (s)