Proof step 1 from proof exdivmod (conclusion) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 2:

  j ≠ 0 ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j)

Rule argument

Additional information

Proof step 2 from proof exdivmod (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 3:

  ⊦ 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j)

• Proof step 4:

  j ≠ 0, 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j) ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j)

Rule argument

Additional information

Used terms: i, j

Proof step 3 from proof exdivmod (previous step) (back to tree)

Lemma exdivmodpos

Proof step 4 from proof exdivmod (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  j ≠ 0, 0 ≤ i ∧ 0 < j → i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j)

Rule argument

Additional information

Proof step 5 from proof exdivmod (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 6:

  j ≠ 0, ¬ j < 0, ¬ i < 0, 0 ≤ i ∧ 0 < j → i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j)

• Proof step 8:

  j ≠ 0, 0 ≤ i ∧ 0 < j → i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j), ¬ i < 0 ∧ ¬ j < 0

Rule argument

Additional information

Used formulas: ¬ i < 0 ∧ ¬ j < 0

Proof step 6 from proof exdivmod (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  j0 < j, 0 < j, 0 ≤ i0 * j + j0, 0 ≤ i0, 0 ≤ j0 ⊦ ∃ i1, j1. i0 * j + j0 = i1 * j + j1 ∧ abs(j1) < j

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s) forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs ⊦ 0 < j → abs(j) = j; (s) ⊦ i < j ↔ ¬ j ≤ i; (ss) ⊦ i < j → ¬ j < i; (s) ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)

Proof step 7 from proof exdivmod (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ i → abs(i) = i; (s) Used terms: i0, j0

Proof step 8 from proof exdivmod (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  j ≠ 0, 0 ≤ i ∧ 0 < j → i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j), ¬ i < 0 ∧ ¬ j < 0

Rule argument

Additional information

Proof step 9 from proof exdivmod (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 10:

  j ≠ 0, j < 0, 0 ≤ i ∧ 0 < j → i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j), ¬ i < 0 ∧ ¬ j < 0

• Proof step 23:

  j ≠ 0, ¬ j < 0, 0 ≤ i ∧ 0 < j → i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j), ¬ i < 0 ∧ ¬ j < 0

Rule argument

Additional information

Used formulas: j < 0

Proof step 10 from proof exdivmod (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  j ≠ 0, j < 0 ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) 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) ⊦ i < j → ¬ j < i; (s)

Proof step 11 from proof exdivmod (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 12:

  ⊦ 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j)

• Proof step 13:

  j ≠ 0, j < 0, 0 ≤ ~ i ∧ 0 < ~ j → (∃ i0, j0. ~ i = i0 * ~ j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < ~ j) ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Rule argument

Additional information

Used terms: ~ i, ~ j

Proof step 12 from proof exdivmod (previous step) (back to tree)

Lemma exdivmodpos

Proof step 13 from proof exdivmod (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  j ≠ 0, j < 0, i ≤ 0 → i + j0 = j * i0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 + j < 0 ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Rule argument

Additional information

Used simplifier rules: minus-left: ⊦ ~ i * j = ~(i * j); associativity: ⊦ i * j * j0 = (i * j) * j0; commutativity: ⊦ i * j = j * i; ⊦ i < ~ j ↔ i + j < 0; (s) add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 ⊦ 0 + i = i; (s) ⊦ i - j = i0 ↔ i = i0 + j; (s) ⊦ i + (j - i0) = i + j - i0; (s) ⊦ ~ i = j ↔ i + j = 0; (s) ⊦ i + ~ j = i - j; (s) ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) ⊦ 0 < ~ i ↔ i < 0; (s) ⊦ i ≤ ~ j ↔ i + j ≤ 0; (s)

Proof step 14 from proof exdivmod (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 15:

  j ≠ 0, j < 0, ¬ 0 < i, i ≤ 0 → i + j0 = j * i0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 + j < 0 ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

• Proof step 17:

  j ≠ 0, 0 < i, j < 0, i ≤ 0 → i + j0 = j * i0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 + j < 0 ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Rule argument

Additional information

Used formulas: ¬ 0 < i

Proof step 15 from proof exdivmod (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  i + j0 = j * i0, j ≠ 0, j0 + j < 0, j < 0, 0 ≤ j0, 0 ≤ i0, i ≤ 0 ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s) add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 ⊦ i + j < j ↔ i < 0; (s) forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs ⊦ i < j ↔ ¬ j ≤ i; (ss)

Proof step 16 from proof exdivmod (previous step) (back to tree)

User applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; ⊦ i + ~ j = i - j; (s) add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 add-c: ⊦ i + j = j + i; from specification int-basic2 ⊦ i - j = i0 ↔ i = i0 + j; (s) ⊦ abs(~ i) = abs(i); (s) ⊦ 0 ≤ i → abs(i) = i; (s) ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) Used terms: i0, ~ j0

Proof step 17 from proof exdivmod (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  j ≠ 0, j < i, j < 0, 0 < i, j + 1 < i ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ↔ ¬ j < i; (ss) ⊦ i < j → (j ≤ i ↔ false); (as) ⊦ i +1 = i + 1; (s) trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1 lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1 forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs

Proof step 18 from proof exdivmod (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 19:

  ⊦ 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j)

• Proof step 20:

  j ≠ 0, j + 1 < i, 0 < i, j < 0, j < i, 0 ≤ i ∧ 0 < ~ j → (∃ i0, j0. i = i0 * ~ j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < ~ j) ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Rule argument

Additional information

Used terms: i, ~ j

Proof step 19 from proof exdivmod (previous step) (back to tree)

Lemma exdivmodpos

Proof step 20 from proof exdivmod (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 21:

  j ≠ 0, j < i, j < 0, 0 < i, j + 1 < i, ∃ i0, j0. i = i0 * ~ j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < ~ j ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < ~ i ↔ i < 0; (s) ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) ⊦ i ≤ j ↔ ¬ j < i; (ss) ⊦ i < j → ¬ j < i; (s)

Proof step 21 from proof exdivmod (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  j ≠ 0, j0 < ~ j, j + 1 < i0 * ~ j + j0, 0 < i0 * ~ j + j0, j < 0, j < i0 * ~ j + j0, 0 ≤ i0, 0 ≤ j0 ⊦ ∃ i1, j1. i0 * ~ j + j0 = i1 * j + j1 ∧ abs(j1) + j < 0

Rule argument

Additional information

Proof step 22 from proof exdivmod (previous step) (back to tree)

User applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: minus-left: ⊦ ~ i * j = ~(i * j); associativity: ⊦ i * j * j0 = (i * j) * j0; commutativity: ⊦ i * j = j * i; ⊦ i < ~ j ↔ i + j < 0; (s) ⊦ 0 + i = i; (s) ⊦ i < j - i0 ↔ i + i0 < j; (s) ⊦ i + ~ j = i - j; (s) ⊦ i - j = i0 ↔ i = i0 + j; (s) ⊦ i + (j - i) = j; (s) add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 add-c: ⊦ i + j = j + i; from specification int-basic2 ⊦ 0 ≤ i → abs(i) = i; (s) ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) Used terms: ~ i0, j0

Proof step 23 from proof exdivmod (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  i < 0, 0 < j, ¬ 0 ≤ i ⊦ ∃ i1, j1. i = i1 * j + j1 ∧ abs(j1) < j

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs ⊦ 0 < j → abs(j) = j; (s) ⊦ i < j → ¬ j < i; (s) ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)

Proof step 24 from proof exdivmod (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 25:

  ⊦ 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j)

• Proof step 26:

  0 < j, i < 0, ¬ 0 ≤ i, 0 ≤ ~ i ∧ 0 < j → (∃ i0, j0. ~ i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j) ⊦ ∃ i1, j1. i = i1 * j + j1 ∧ abs(j1) < j

Rule argument

Additional information

Used terms: ~ i, j

Proof step 25 from proof exdivmod (previous step) (back to tree)

Lemma exdivmodpos

Proof step 26 from proof exdivmod (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  i + i0 * j + j0 = 0, j0 < j, i < 0, 0 < j, 0 ≤ j0, 0 ≤ i0, i ≤ 0 ⊦ ∃ i1, j1. i = i1 * j + j1 ∧ abs(j1) < j

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs ⊦ i < j → ¬ j < i; (s) ⊦ i ≤ j ↔ ¬ j < i; (ss) ⊦ ~ i = j ↔ i + j = 0; (s) add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 ⊦ i + 0 = i; (s) ⊦ i ≤ ~ j ↔ i + j ≤ 0; (s)

Proof step 27 from proof exdivmod (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 28:

  i + i0 * j + j0 = 0, 0 < j, i < 0, j0 < j, i ≤ 0, 0 ≤ i0, 0 ≤ j0 ⊦ ∃ i1, j1. i = i1 * j + j1 ∧ abs(j1) < j, i = (~ i0 - 1) * j + (j - j0) ∧ abs(j - j0) < j

Rule argument

Additional information

Used terms: ~ i0 - 1, j - j0

Proof step 28 from proof exdivmod (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  i + i0 * j + j0 = 0, j0 < j, i < 0, 0 < j, 0 ≤ j0, 0 ≤ i0, i ≤ 0 ⊦ ∃ i1, j1. i = i1 * j + j1 ∧ abs(j1) < j, i = j - j0 - i0 * j + j ∧ j - j0 < j

Rule argument

Additional information

Used simplifier rules: minus-left: ⊦ ~ i * j = ~(i * j); dist-lsplit: ⊦ (i + i0) * j = i * j + i0 * j; one: ⊦ 1 * i = i; associativity: ⊦ i * j * j0 = (i * j) * j0; commutativity: ⊦ i * j = j * i; ⊦ j < i → abs(i - j) = i - j; (s) add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 ⊦ i + ~ j = i - j; (s) ⊦ ~ i - j = ~(i + j); (s)

Proof step 29 from proof exdivmod (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 30:

  i + i0 * j + j0 = 0, 0 < j, i < 0, j0 < j, i ≤ 0, 0 ≤ i0, 0 ≤ j0 ⊦ i = j - j0 - i0 * j + j ∧ j - j0 < j, i = ~ i0 * j + 0 ∧ abs(0) < j

Rule argument

Additional information

Used terms: ~ i0, 0

Proof step 30 from proof exdivmod (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: minus-left: ⊦ ~ i * j = ~(i * j); associativity: ⊦ i * j * j0 = (i * j) * j0; commutativity: ⊦ i * j = j * i; add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 add-c: ⊦ i + j = j + i; from specification int-basic2 ⊦ 0 + i = i; (s) trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1 lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1 ⊦ i ≤ j → (j ≤ i ↔ i = j); (s) ⊦ i < j ↔ ¬ j ≤ i; (ss) ⊦ i < i + j ↔ 0 < j; (s) ⊦ i - j < i0 ↔ i < j + i0; (s) ⊦ ~ i = j ↔ i + j = 0; (s) ⊦ i - j + i = ~ j; (s) ⊦ i - j - i0 = i - j + i0; (s) ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s)