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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  m0 * n + n0 + 1 = m * n + n1, n1 ≠ 0, n1 < n0, n0 < n ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  m0 * n + n0 + 1 = m * n + n1, n1 ≠ 0, n1 < n, n1 < n0, n0 < n, n1 + 1 < n ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 4:

  ⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1)

• Proof step 5:

  m0 * n + n0 + 1 = m * n + n1, n1 ≠ 0, n1 + 1 < n, n0 < n, n1 < n0, n1 < n, n0 + 1 < n ∧ n1 < n → (m0 * n + n0 + 1 = m * n + n1 ↔ m0 = m ∧ n0 + 1 = n1) ⊦

Rule argument

Additional information

Used terms: n0 + 1, n, m0, m, n1

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

Lemma modeq

Proof step 5 from proof mod-neq-03 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  (n0 * m0 + m0) + n0 + 1 = (n0 * m + m) + n1, n1 ≠ 0, n1 < n0 ⊦

Rule argument

Additional information

Used simplifier rules: modeq: ⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1); rsucc-01: ⊦ (m + 1) * n = m * n + n; mult-a: ⊦ m0 * m * n = (m0 * m) * n; mult-c: ⊦ m * n = n * m; c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m < n → ¬ n < m; (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s) ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ¬ m + n < m; (s) ⊦ m < n → m ≤ n; (s) lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic

Proof step 6 from proof mod-neq-03 (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 7:

  ⊦ n0 < n ∧ n0 ≠ 0 → m * n ≠ m0 * n + n0

Rule argument

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n; mult-a: ⊦ m0 * m * n = (m0 * m) * n; mult-c: ⊦ m * n = n * m; ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m < n → ¬ n < m; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic Used terms: n1, n0 + 1, m0 + 1, m

Proof step 7 from proof mod-neq-03 (previous step) (back to tree)

Lemma mod-neq-02