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Proof step 16 from proof aligndown-is-n (conclusion)

n ≠ 0 ⊦ n ≤ n0 + n1 ∧ n0 + n1 < 2 * n → align↓(n0 + n1, n) = 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-mult,,ls-04, ⊦ m0 ≠ 0 → (m0 < m0 * m ↔ 1 < m)) Csimpnamedseq(nat-mult,,mult-c, ⊦ m * n = n * m) Csimpnamedseq(nat-mult,,mult-a, ⊦ m0 * m * n = (m0 * m) * n) Csimpnamedseq(nat-mult,,two-zero-01, ⊦ 2 * m = 0 ↔ m = 0) Csimpnamedforward(nat,,fle-01, ⊦ m ≤ n ∧ n < n0 → m < n0) Csimpnamedforward(nat-basic1,,f, ⊦ m < n → n ≠ 0)))]