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Proof step 3 from proof is-aligned-minus (conclusion)

m ≠ 0 ⊦ n1 ≤ n0 ∧ is-aligned(n0, m) ∧ is-aligned(n1, m) → is-aligned(n0 - n1, m)

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)))]