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Proof step 9 from proof aligndown-is-zero (conclusion)

n ≠ 0 ⊦ align↓(m, n) = 0 ↔ m < n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
0 ≠ m ⊦ align↓(n, m) = (n % m = 0 ⊃ n;n / m * m)

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(aligndown-def,,, 0 ≠ m ⊦ align↓(n, m) = (n % m = 0 ⊃ n;n / m * m)) Csimpnamedseq(nat,,lels, ⊦ ¬ m < n ↔ n ≤ m) Csimpnamedseq(nat-mult,,mult-c, ⊦ m * n = n * m) Csimpnamedseq(nat-mult,,mult-a, ⊦ m0 * m * n = (m0 * m) * n) Csimpnamedseq(nat-mult,,mult-base, ⊦ m * 0 = 0) Csimpnamedseq(nat-div,,zero-01, ⊦ m < n → m / n = 0) Csimpnamedseq(nat-div,,mod-eq, ⊦ m < n → m % n = m)))]