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Proof step 2 from proof between-is-aligned (previous step)

m0 < m1, m1 < m0 + m, is-aligned(m1, m), is-aligned(m0, m) ⊦ m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ is-aligned(n, m) ↔ n % m = 0
Proof step 4:
m1 % m = 0, m0 % m = 0, m0 < m1, m1 < m0 + m ⊦ m = 0, m = 1

Rule argument

Additional information

Used simplifier rules: is-aligned-def: ⊦ is-aligned(n, m) ↔ n % m = 0;% ⊦ m < m + n ↔ n ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic