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

is-aligned(n1, m) ⊦ is-aligned(n0 + n1, m) ↔ is-aligned(n0, 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:
m ≠ 0 ⊦ is-aligned(n, m) → n % m = 0

Rule argument

Additional information

Used simplifier rules: is-aligned-def: ⊦ is-aligned(n, m) ↔ n % m = 0;%is-aligned-mod: m ≠ 0 ⊦ is-aligned(n, m) → n % m = 0;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m0 % n = 0 ∧ 0 ≠ n → (m + m0) % n = m % n; (s)