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Proof step 9 from proof alignup-add-bounded1 (previous step)

n ≤ m ⊦ n % m = 0, n / m = 0, m = 0, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ m ≤ n → (n ≤ m ↔ m = n)

Rule argument

spec: nat name: anti seq: ⊦ m ≤ n → (n ≤ m ↔ m = n) subst: [m, n] → [m, n] flags: <all paths>

Additional information

Used simplifier rules: n ≠ 0 ⊦ n / n = 1; (s)%n ≠ 0 ⊦ n % n = 0; (s)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m < n → m / n = 0; (s) Used terms: m, n