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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦

Rule argument

Additional information

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 4:

  0 ≤ i, 0 ≤ j ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j)

• Proof step 5:

  ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦ 0 ≤ i ∧ 0 ≤ j

Rule argument

Additional information

Used terms: i, j

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

Lemma labsp

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦ 0 ≤ i ∧ 0 ≤ j

Rule argument

Additional information

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 7:

  0 ≤ i, 0 ≤ j ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j)

• Proof step 8:

  ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦ 0 ≤ i ∧ 0 ≤ j, 0 ≤ i ∧ 0 ≤ ~ j

• Proof step 18:

  i→n(i) < i→n(~ j) ↔ abs(i) < abs(~ j), ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦ 0 ≤ i ∧ 0 ≤ j

Rule argument

Additional information

Used terms: i, ~ j

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

Lemma labsp

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦ 0 ≤ i ∧ 0 ≤ j, 0 ≤ i ∧ j ≤ 0

Rule argument

Additional information

Used simplifier rules: 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 9 from proof labs (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 10:

  0 ≤ i, 0 ≤ j ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j)

• Proof step 11:

  ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦ 0 ≤ i ∧ j ≤ 0, 0 ≤ i ∧ 0 ≤ j, 0 ≤ ~ i ∧ 0 ≤ j

• Proof step 17:

  i→n(~ i) < i→n(j) ↔ abs(~ i) < abs(j), ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦ 0 ≤ i ∧ j ≤ 0, 0 ≤ i ∧ 0 ≤ j

Rule argument

Additional information

Used terms: ~ i, j

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

Lemma labsp

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦ 0 ≤ i ∧ 0 ≤ j, 0 ≤ i ∧ j ≤ 0, i ≤ 0 ∧ 0 ≤ j

Rule argument

Additional information

Used simplifier rules: 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 12 from proof labs (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 13:

  0 ≤ i, 0 ≤ j ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j)

• Proof step 14:

  ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦ i ≤ 0 ∧ 0 ≤ j, 0 ≤ i ∧ j ≤ 0, 0 ≤ i ∧ 0 ≤ j, 0 ≤ ~ i ∧ 0 ≤ ~ j

• Proof step 16:

  i→n(~ i) < i→n(~ j) ↔ abs(~ i) < abs(~ j), ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦ i ≤ 0 ∧ 0 ≤ j, 0 ≤ i ∧ j ≤ 0, 0 ≤ i ∧ 0 ≤ j

Rule argument

Additional information

Used terms: ~ i, ~ j

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

Lemma labsp

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  ¬(i→n(i) < i→n(j) ↔ abs(i) < abs(j)) ⊦ 0 ≤ i ∧ 0 ≤ j, 0 ≤ i ∧ j ≤ 0, i ≤ 0 ∧ 0 ≤ j, i ≤ 0 ∧ j ≤ 0

Rule argument

Additional information

Used simplifier rules: 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 15 from proof labs (previous step) (back to tree)

User applied rule case distinction

And closed the goal!

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i); ⊦ abs(~ i) = abs(i); (s)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i); ⊦ abs(~ i) = abs(i); (s)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i); ⊦ abs(~ i) = abs(i); (s)