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Proof step 4 from proof alignup-div-mult (previous step)

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: n ≠ 0 ⊦ (n0 + m * n) / n = n0 / n + m; (s)%n ≠ 0 ⊦ n / n = 1; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m * (n + 1) = m * n + m; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ n % m = 0 ∧ m ≠ 0 → m * n / m = n; (s)