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Proof step 1 from proof numeral-04 (conclusion) (back to tree)

m2 < m1 ⊦ (* m2) * m + n ≤ (* m1) * m + n0 ↔ n ≤ (m1 - m2) * m + n0

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof numeral-04 (previous step) (back to tree)

¬(m2 * m + n ≤ m1 * m + n0 ↔ n ≤ (m1 - m2) * m + n0), m2 < m1 ⊦

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 numeral-04 (previous step) (back to tree)

¬(m2 * m + n ≤ m1 * m + n0 ↔ n ≤ (m1 - m2) * m + n0), m2 < m1 ⊦

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

spec: nat, name:elim-sub, seq:n ≤ m ⊦ n0 = m - n ↔ m = n0 + n, subst:[m, n, n0] → [m1, m2, n1]

Additional information

Used simplifier rules: dist: ⊦ m0 * (m + n) = m0 * m + m0 * n; mult-a: ⊦ m0 * m * n = (m0 * m) * n; mult-c: ⊦ m * n = n * m; ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m < n → m ≤ n; (s) n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m1, m2, n1

Proof Tree for numeral-04 (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