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Proof step 4 from proof between-is-aligned (conclusion)

m ≠ 0 ⊦ is-aligned(m0, m) ∧ m0 < m1 ∧ m1 < m0 + m → (is-aligned(m1, m) ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ is-aligned(n, m) ↔ n % m = 0

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(is-aligned-def,,, ⊦ is-aligned(n, m) ↔ n % m = 0) Csimpnamedseq(nat-basic2,,l-01, ⊦ m < m + n ↔ n ≠ 0) Csimpnamedseq(nat-basic2,,c, ⊦ m + n = n + m) Csimpnamedseq(nat-basic2,,a, ⊦ (m + n0) + n1 = m + n0 + n1) Csimpnamedseq(nat-basic1,,one, ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1)) Csimpnamedseq(nat-basic2,,l, ⊦ m + n < m + n0 ↔ n < n0) Csimpnamedseq(nat-basic2,,succ, ⊦ m +1 = m + 1) Csimpnamedforward(nat-basic1,,trans-1, ⊦ n < n0 ∧ n0 < n1 → n < n1) Csimpnamedforward(nat-basic1,,lf, ⊦ m < n ∧ n < n0 → m +1 < n0) Csimpnamedforward(nat-basic1,,f, ⊦ m < n → n ≠ 0)))]