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

⊦ (n0 % m = 0 ⊃ n * m + n0;m * (n0 / m + n) + m) = n * m + (n0 % m = 0 ⊃ n0;m * n0 / 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: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ m * n + m0 * n = m1 * n ↔ m + m0 = m1 ∨ 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%c: ⊦ m + n = n + m; from specification nat-basic