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Proof step 1 from proof mod-ls-03 (conclusion) (back to tree)

n ≠ 0 ⊦ ¬ n < m % n

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof mod-ls-03 (previous step) (back to tree)

n ≠ 0, n < m % n ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 3 from proof mod-ls-03 (previous step) (back to tree)

n ≠ 0, n < m % n ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:div-elim, seq:n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n, subst:[m, n, m0, n0] → [m, n, m0, n0]

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) ⊦ m < n → ¬ n < m; (s) Used terms: m, n, m0, n0

Proof step 4 from proof mod-ls-03 (previous step) (back to tree)

n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n

Lemma div-elim

Proof Tree for mod-ls-03 (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.

the proof tree