Proof step 1 from proof div-abs-02 (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(i | abs(j) ↔ i | j) ⊦

Rule argument

Additional information

Proof step 2 from proof div-abs-02 (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 3:

  i | abs(j), ¬ i | j ⊦

• Proof step 14:

  i | j, ¬ i | abs(j) ⊦

Rule argument

Additional information

Proof step 3 from proof div-abs-02 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ i | j ↔ (∃ i0. i * i0 = j)

• Proof step 5:

  ¬ i | j, ∃ i0. i * i0 = abs(j) ⊦

Rule argument

Additional information

Used terms: abs(j), i

Proof step 4 from proof div-abs-02 (previous step) (back to tree)

Lemma divides

Proof step 5 from proof div-abs-02 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  i * i0 = abs(j), ¬ i | j ⊦

Rule argument

Additional information

Proof step 6 from proof div-abs-02 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ i | j ↔ (∃ i0. i * i0 = j)

• Proof step 8:

  i * i0 = abs(j), ∀ i0. i * i0 ≠ j ⊦

Rule argument

Additional information

Used terms: j, i

Proof step 7 from proof div-abs-02 (previous step) (back to tree)

Lemma divides

Proof step 8 from proof div-abs-02 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

  ⊦ ¬ j < 0 → abs(j) = j

• Proof step 10:

  i * i0 = abs(j), j < 0, ∀ i0. i * i0 ≠ j ⊦

• Proof step 12:

  i * i0 = j, ¬ j < 0, ∀ i0. i * i0 ≠ j ⊦

Rule argument

Additional information

Used terms: j

Proof step 9 from proof div-abs-02 (previous step) (back to tree)

A spec-lemma

Proof step 10 from proof div-abs-02 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  j + i * i0 = 0, j < 0, ∀ i0. i * i0 ≠ j ⊦

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs ⊦ ~ i = j ↔ i + j = 0; (s) ⊦ i < 0 → abs(i) = ~ i; (s)

Proof step 11 from proof div-abs-02 (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i * j = ~(i * j); (s) associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult commutativity: ⊦ i * j = j * i; from specification int-mult ⊦ ~ i = j ↔ i + j = 0; (s) add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: ~ i0

Proof step 12 from proof div-abs-02 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  0 ≤ i * i0, ∀ i1. i * i1 ≠ i * i0 ⊦

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs ⊦ i < j ↔ ¬ j ≤ i; (ss)

Proof step 13 from proof div-abs-02 (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: i0

Proof step 14 from proof div-abs-02 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 15:

  ⊦ i | j ↔ (∃ i0. i * i0 = j)

Rule argument

Additional information

Used simplifier rules: div-abs: ⊦ i | abs(i) * j; ⊦ abs(i * j) = abs(i) * abs(j); (s) associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j, i

Proof step 15 from proof div-abs-02 (previous step) (back to tree)

Lemma divides