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Proof step 1 from proof square-ex (conclusion) (back to tree)

⊦ 0 < i → (∃ j. 0 < j ∧ j * j ≤ i ∧ i < (j + 1) * (j + 1))

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof square-ex (previous step) (back to tree)

0 < i ⊦ ∃ j. 0 < j ∧ j * j ≤ i ∧ i < (j + 1) * (j + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs

Proof step 3 from proof square-ex (previous step) (back to tree)

0 < i ⊦ ∃ j. 0 < j ∧ j * j ≤ i ∧ i < (j + 1) * (j + 1)

User applied rule structural induction

And obtained the following new goals

Rule argument

indvar: i

Additional information

Used simplifier rules: int generated by 0, +1, -1;;

Proof step 4 from proof square-ex (previous step) (back to tree)

0 < i0 +1, 0 < i0 → (∃ j. 0 < j ∧ j * j ≤ i0 ∧ i0 < (j + 1) * (j + 1)) ⊦ ∃ j. 0 < j ∧ j * j ≤ i0 +1 ∧ i0 +1 < (j + 1) * (j + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 5 from proof square-ex (previous step) (back to tree)

0 ≤ i0, 0 < i0 → 0 < j ∧ j * j ≤ i0 ∧ i0 < (j + 1) * (j + 1) ⊦ ∃ j0. 0 < j0 ∧ j0 * j0 ≤ i0 + 1 ∧ i0 + 1 < (j0 + 1) * (j0 + 1)

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: 0 < i0

Additional information

Used formulas: 0 < i0

Proof step 6 from proof square-ex (previous step) (back to tree)

0 < i0, 0 ≤ i0, 0 < i0 → 0 < j ∧ j * j ≤ i0 ∧ i0 < (j + 1) * (j + 1) ⊦ ∃ j0. 0 < j0 ∧ j0 * j0 ≤ i0 + 1 ∧ i0 + 1 < (j0 + 1) * (j0 + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: square-ls: ⊦ i * i < j * j ↔ abs(i) < abs(j); associativity: ⊦ i * j * j0 = (i * j) * j0; commutativity: ⊦ i * j = j * i; add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 ⊦ i ≤ i; (s) ⊦ i < j + 1 ↔ i ≤ j; (s) ⊦ 0 < j → abs(j + 1) = j + 1; (s) ⊦ 0 < j → abs(j) = j; (s) forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1

Proof step 7 from proof square-ex (previous step) (back to tree)

1 < (j + 1) * (j + 1), 0 < (j + 1) * (j + 1), i0 < (j + 1) * (j + 1), 0 < j, 0 < i0, j * j ≤ i0, 0 ≤ i0 ⊦ ∃ j0. 0 < j0 ∧ j0 * j0 ≤ i0 + 1 ∧ i0 + 1 < (j0 + 1) * (j0 + 1)

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: j abort: F discard: F computed: F

Additional information

Used terms: j

Proof step 8 from proof square-ex (previous step) (back to tree)

0 < i0, 0 < j, i0 < (j + 1) * (j + 1), 0 < (j + 1) * (j + 1), 1 < (j + 1) * (j + 1), 0 ≤ i0, j * j ≤ i0 ⊦ ∃ j0. 0 < j0 ∧ j0 * j0 ≤ i0 + 1 ∧ i0 + 1 < (j0 + 1) * (j0 + 1), 0 < j ∧ j * j ≤ i0 + 1 ∧ i0 + 1 < (j + 1) * (j + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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 ≤ i; (s) ⊦ i < j + 1 ↔ i ≤ j; (s) ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s) ⊦ ~1 < i ↔ 0 ≤ i; (s) ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s) forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1 ⊦ i < j → (j ≤ i + 1 ↔ j = i + 1); (s) ⊦ i < j ↔ ¬ j ≤ i; (ss) ⊦ i ≤ i0 ∧ j0 ≥ 0 → i ≤ i0 + j0; (s)

Proof step 9 from proof square-ex (previous step) (back to tree)

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

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: j + 1 abort: F discard: T computed: F

Additional information

Used terms: j + 1

Proof step 10 from proof square-ex (previous step) (back to tree)

(j + 1) * (j + 1) = i0 + 1, 0 < i0, 0 < j, j * j ≤ i0, 0 ≤ i0 ⊦ 0 < j + 1 ∧ (j + 1) * (j + 1) ≤ i0 + 1 ∧ i0 + 1 < ((j + 1) + 1) * ((j + 1) + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 11 from proof square-ex (previous step) (back to tree)

(j + 1) * (j + 1) = i0 + 1, 0 < i0, 0 < j, j * j ≤ i0, 0 ≤ i0, (2 + j) * (2 + j) ≤ i0 + 1 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:dist-lsplit, seq: ⊦ (i + i0) * j = i * j + i0 * j, subst:[i, i0, j] → [2, j, 2 + j], <all paths>

Additional information

Used terms: 2, j, 2 + j

Proof step 12 from proof square-ex (previous step) (back to tree)

⊦ (i + i0) * j = i * j + i0 * j

Lemma dist-lsplit

Proof step 13 from proof square-ex (previous step) (back to tree)

(j + 1) * (j + 1) = i0 + 1, 0 < j, 0 < i0, 2 * (2 + j) + j * (2 + j) ≤ i0 + 1, 0 ≤ i0, j * j ≤ i0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:dist-lsplit, seq: ⊦ (i + i0) * j = i * j + i0 * j, subst:[i, i0, j] → [j, 1, j + 1], <all paths>

Additional information

Used simplifier rules: le-dist-02: ⊦ i * (j + i0) + j2 ≤ i * (j0 + i0) + j1 ↔ i * j + j2 ≤ i * j0 + j1; numle-03: ⊦ (* i2) * i + j ≤ i + j0 ↔ (i2 - 1) * i + j ≤ j0; one: ⊦ 1 * i = i; le-one: ⊦ 1 < j → (i * j ≤ 1 ↔ i ≤ 0); associativity: ⊦ i * j * j0 = (i * j) * j0; commutativity: ⊦ i * j = j * i; ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); (s) ⊦ 0 + i = i; (s) ⊦ i + j ≤ i + j0 ↔ j ≤ j0; (s) ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; (s) ⊦ i < 0 ∧ 0 < j → ¬ j ≤ i; (s) add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j, 1, j + 1

Proof step 14 from proof square-ex (previous step) (back to tree)

⊦ (i + i0) * j = i * j + i0 * j

Lemma dist-lsplit

Proof step 15 from proof square-ex (previous step) (back to tree)

0 ≤ i0, ¬ 0 < i0, 0 < i0 → 0 < j ∧ j * j ≤ i0 ∧ i0 < (j + 1) * (j + 1) ⊦ ∃ j0. 0 < j0 ∧ j0 * j0 ≤ i0 + 1 ∧ i0 + 1 < (j0 + 1) * (j0 + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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 ≤ j → (j ≤ i ↔ i = j); (s) ⊦ i < j ↔ ¬ j ≤ i; (ss)

Proof step 16 from proof square-ex (previous step) (back to tree)

⊦ ∃ j0. 0 < j0 ∧ j0 * j0 ≤ 1 ∧ 1 < (j0 + 1) * (j0 + 1)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: 1 abort: F discard: F computed: F

Additional information

Used simplifier rules: 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 Used terms: 1

Proof step 17 from proof square-ex (previous step) (back to tree)

0 < i0 -1, 0 < i0 → (∃ j. 0 < j ∧ j * j ≤ i0 ∧ i0 < (j + 1) * (j + 1)) ⊦ ∃ j. 0 < j ∧ j * j ≤ i0 -1 ∧ i0 -1 < (j + 1) * (j + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: square-ls: ⊦ i * i < j * j ↔ abs(i) < abs(j); associativity: ⊦ i * j * j0 = (i * j) * j0; commutativity: ⊦ i * j = j * i; 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 ≤ i; (s) ⊦ i < j + 1 ↔ i ≤ j; (s) ⊦ 0 < j → abs(j + 1) = j + 1; (s) ⊦ 0 < j → abs(j) = j; (s) forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1 ⊦ i - j < i0 ↔ i < j + i0; (s) ⊦ i -1 = i - 1; (s) ⊦ i + 1 ≤ j ↔ i < j; (s) ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s) ⊦ i < j - i0 ↔ i + i0 < j; (s)

Proof step 18 from proof square-ex (previous step) (back to tree)

1 < (j + 1) * (j + 1), i0 < (j + 1) * (j + 1), 0 < j, 0 < i0, 1 < i0, 0 < (j + 1) * (j + 1), j * j ≤ i0 ⊦ ∃ j0. 0 < j0 ∧ j0 * j0 < i0 ∧ i0 ≤ (j0 + 1) * (j0 + 1)

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: j abort: F discard: F computed: F

Additional information

Used terms: j

Proof step 19 from proof square-ex (previous step) (back to tree)

0 < (j + 1) * (j + 1), 1 < i0, 0 < i0, 0 < j, i0 < (j + 1) * (j + 1), 1 < (j + 1) * (j + 1), j * j ≤ i0 ⊦ ∃ j0. 0 < j0 ∧ j0 * j0 < i0 ∧ i0 ≤ (j0 + 1) * (j0 + 1), 0 < j ∧ j * j < i0 ∧ i0 ≤ (j + 1) * (j + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: square-ls: ⊦ i * i < j * j ↔ abs(i) < abs(j); square-le: ⊦ i * i ≤ j * j ↔ abs(i) ≤ abs(j); zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j); associativity: ⊦ i * j * j0 = (i * j) * j0; commutativity: ⊦ i * j = j * i; forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1 add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 ⊦ i ≤ i; (s) ⊦ i < j + 1 ↔ i ≤ j; (s) ⊦ i < j → (j ≤ i + 1 ↔ j = i + 1); (s) ⊦ 0 < j → abs(j + 1) = j + 1; (s) ⊦ 0 < j → abs(j) = j; (s) ⊦ i ≤ i + j ↔ 0 ≤ j; (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 ⊦ i ≤ j → (j ≤ i ↔ i = j); (s) ⊦ i < j ↔ ¬ j ≤ i; (ss) ⊦ i < j → ¬ j < i; (s) ⊦ i ≤ j ↔ ¬ j < i; (ss)

Proof step 20 from proof square-ex (previous step) (back to tree)

1 < (j + 1) * (j + 1), 0 < (j + 1) * (j + 1), 1 < j * j, 0 < j ⊦ ∃ j0. 0 < j0 ∧ j = j0 + 1

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: j -1 abort: F discard: F computed: T

Additional information

Used simplifier rules: one-01: ⊦ i = 1 → i * j = j; associativity: ⊦ i * j * j0 = (i * j) * j0; commutativity: ⊦ i * j = j * i; ⊦ i < j - i0 ↔ i + i0 < j; (s) ⊦ i -1 = i - 1; (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 ⊦ i < j ↔ ¬ j ≤ i; (ss) ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s) forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1 Used terms: j -1

Proof Tree for square-ex (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.

the proof tree