Proof step 1 from proof div-eq-same-01 (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(i / j = i ↔ i = 0 ∨ j = 1), j ≠ 0 ⊦

Rule argument

Additional information

Proof step 2 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ¬(i / j = i ↔ i = 0 ∨ j = 1), j ≠ 0 ⊦

Rule argument

Additional information

Proof step 3 from proof div-eq-same-01 (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 4:

  ¬(i / j = i ↔ i = 0 ∨ j = 1), i ≠ 0, j ≠ 0 ⊦

Rule argument

Additional information

Used simplifier rules: zero: j ≠ 0 ⊦ 0 / j = 0; Used formulas: i = 0

Proof step 4 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  i / j = i, i ≠ 0, j ≠ 0, j ≠ 1 ⊦

Rule argument

Additional information

Used simplifier rules: one: ⊦ i / 1 = i;

Proof step 5 from proof div-eq-same-01 (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 6:

  ⊦ j < 0 ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ j < j0 ∧ 0 ≤ i0)

• Proof step 7:

  i / j = i, i ≠ 0, j ≠ 0, j ≠ 1 ⊦ j < 0 ∧ i < 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: i, j, i0, j0

Proof step 6 from proof div-eq-same-01 (previous step) (back to tree)

Lemma div-elim-negneg

Proof step 7 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  i / j = i, i ≠ 0, j ≠ 0, j ≠ 1 ⊦ j < 0 ∧ i < 0

Rule argument

Additional information

Proof step 8 from proof div-eq-same-01 (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 9:

  ⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)

• Proof step 10:

  i / j = i, i ≠ 0, j ≠ 0, j ≠ 1 ⊦ j < 0 ∧ i < 0, 0 < j ∧ 0 ≤ i

• Proof step 24:

  i = i0 * j + j0, i0 = i, i ≠ 0, j ≠ 0, j ≠ 1, j0 < j, 0 < j, 0 ≤ j0, 0 ≤ i0, 0 ≤ i ⊦ j < 0 ∧ i < 0

Rule argument

Additional information

Used terms: i, j, i0, j0

Proof step 9 from proof div-eq-same-01 (previous step) (back to tree)

Lemma div-elim-pospos

Proof step 10 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  i / j = i, i ≠ 0, j ≠ 0, j ≠ 1 ⊦ j < 0 ∧ i < 0, 0 < j ∧ 0 < i

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)

Proof step 11 from proof div-eq-same-01 (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 12:

  i / j = i, i ≠ 0, j ≠ 0, j ≠ 1, 0 < j ⊦ 0 < j ∧ 0 < i, j < 0 ∧ i < 0

• Proof step 21:

  i / j = i, i ≠ 0, j ≠ 0, j ≠ 1, ¬ 0 < j ⊦ 0 < j ∧ 0 < i, j < 0 ∧ i < 0

Rule argument

Additional information

Used formulas: 0 < j

Proof step 12 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  i / j = i, j ≠ 1, j ≠ 0, i < j, i < 0, 0 < j, i + 1 < j ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) ⊦ i +1 = i + 1; (s) 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 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; (s) ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)

Proof step 13 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 14:

  ⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)

• Proof step 15:

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

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) Used terms: i, j, i0, j0

Proof step 14 from proof div-eq-same-01 (previous step) (back to tree)

Lemma div-elim-negpos

Proof step 15 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  i = i * j + j0, j ≠ 0, j ≠ 1, j0 < j, i + 1 < j, 0 < j, i < 0, i < j, abs(j0) < j, j0 ≤ 0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s) forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s)

Proof step 16 from proof div-eq-same-01 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 17:

  ⊦ i = (* i1) * i + j0 ↔ 0 = (i1 - 1) * i + j0

• Proof step 18:

  0 = (j - 1) * i + j0, j ≠ 0, j ≠ 1, abs(j0) < j, i < j, i < 0, 0 < j, i + 1 < j, j0 < j, j0 ≤ 0 ⊦

Rule argument

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j0, j, i

Proof step 17 from proof div-eq-same-01 (previous step) (back to tree)

A spec-lemma

Proof step 18 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 19:

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

Rule argument

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, 1, i0

Proof step 19 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

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

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) ⊦ i +1 = i + 1; (s) forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1 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 add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 ⊦ i < j + 1 ↔ i ≤ j; (s) ⊦ 0 + i = i; (s) ⊦ i + (* j) < i0 + (* j0) ↔ i < i0 + (j0 - j); (s) ⊦ i ≠ j → (i ≤ j ↔ i < j); (s) ⊦ ~1 < i ↔ 0 ≤ i; (s) ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s) ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)

Proof step 20 from proof div-eq-same-01 (previous step) (back to tree)

User applied rule cut formula

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j); (s) ⊦ i < 0 → abs(i) = ~ i; (s) ⊦ ~ i ≤ j ↔ 0 ≤ i + j; (s) forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1 lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1 forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s) ⊦ i +1 = i + 1; (s) ⊦ i < i + j ↔ 0 < j; (s) ⊦ 0 < i → (0 < i * j ↔ 0 < j); (s) ⊦ i < j → ¬ j < i; (s) associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult commutativity: ⊦ i * j = j * i; from specification int-mult 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) ⊦ 0 < i → (i * j = 0 ↔ j = 0); (s) ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s) Used formulas: j0 = 0

Proof step 21 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  i / j = i, j ≠ 1, j < i, j < 0, 0 < i, j + 1 < i ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) ⊦ i +1 = i + 1; (s) forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs 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 ⊦ i < j → ¬ j < i; (s) ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)

Proof step 22 from proof div-eq-same-01 (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 23:

  ⊦ j < 0 ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < abs(j) ∧ i0 ≤ 0)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j → (j ≤ i ↔ i = j); (s) ⊦ i < j → (j ≤ i ↔ false); (as) ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) ⊦ i ≤ j ↔ ¬ j < i; (ss) ⊦ i < j → ¬ j < i; (s) Used terms: i, j, i0, j0

Proof step 23 from proof div-eq-same-01 (previous step) (back to tree)

Lemma div-elim-posneg

Proof step 24 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  i = i * j + j0, i ≠ 0, j ≠ 1, j ≠ 0, 0 < i, j0 < j, 0 < j, 0 ≤ j0 ⊦

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 lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1 ⊦ i < j → ¬ j < i; (s) ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)

Proof step 25 from proof div-eq-same-01 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  ⊦ i = (* i1) * i + j0 ↔ 0 = (i1 - 1) * i + j0

• Proof step 27:

  0 = (j - 1) * i + j0, i ≠ 0, j ≠ 1, j ≠ 0, 0 < j, j0 < j, 0 < i, 0 ≤ j0 ⊦

Rule argument

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j0, j, i

Proof step 26 from proof div-eq-same-01 (previous step) (back to tree)

A spec-lemma

Proof step 27 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 28:

  0 = i0 * i + j0, j = i0 + 1, i ≠ 0, j ≠ 1, j ≠ 0, 0 < i, j0 < j, 0 < j, 0 ≤ j0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, 1, i0

Proof step 28 from proof div-eq-same-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  0 = i0 * i + j0, ~1 ≠ i0, 0 ≠ i0, i ≠ 0, 0 < i, 0 < i0, j0 ≤ i0, 0 ≤ j0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j); (s) forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 ⊦ i < j + 1 ↔ i ≤ j; (s) ⊦ ~1 < i ↔ 0 ≤ i; (s) ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s) ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)

Proof step 29 from proof div-eq-same-01 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 30:

  ⊦ 0 < i ∧ 0 < j → i + j ≠ 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < i → (0 < i * j ↔ 0 < j); (s) ⊦ i < j ↔ ¬ j ≤ i; (ss) ⊦ i ≤ j → (j ≤ i ↔ i = j); (s) ⊦ 0 + i = i; (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 → (i * j = 0 ↔ j = 0); (s) associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j0, i0 * i

Proof step 30 from proof div-eq-same-01 (previous step) (back to tree)

A spec-lemma