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

⊦ 1 < i ∧ 0 < j → ¬ i ^ j ≤ 1

System applied rule normalize

And obtained the following new goals

Proof step 2:
1 < i, 0 < j, i ^ j ≤ 1 ⊦

Rule argument

Additional information

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

1 < i, 0 < j, i ^ j ≤ 1 ⊦

User applied rule induction

And obtained the following new goals

Proof step 3:
1 < i, 0 < j, i ^ j ≤ 1, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Rule argument

indvar: i→n(j) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j)

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

1 < i, 0 < j, i ^ j ≤ 1, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i ^ j < i, 0 < j, 1 < i, i ^ j ≤ 1, 2 ≤ i, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Rule argument

Additional information

Used simplifier rules: numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs ⊦ 1 ≤ i ↔ 0 < i; (s) forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs

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

i ^ j < i, 0 < j, 1 < i, i ^ j ≤ 1, 2 ≤ i, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < i → j ^ i = j ^ i -1 * j
Proof step 6:
1 < i, i ^ j -1 * i < i, 0 < j, 2 ≤ i, i ^ j -1 * i ≤ 1, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Rule argument

spec: <toplevel>, name:Pot-pos, seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j, subst:[i, j] → [j, i], <all paths>

Additional information

Used terms: j, i

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

⊦ 0 < i → j ^ i = j ^ i -1 * j

Lemma Pot-pos

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

1 < i, i ^ j -1 * i < i, 0 < j, 2 ≤ i, i ^ j -1 * i ≤ 1, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i ^ j -1 < i, i ^ j -1 < j, i ^ j -1 < 1, 0 < j, 1 < i, i ^ j -1 + 1 < i, 2 ≤ i, i ^ j -1 ≤ 0, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Rule argument

Additional information

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

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

i ^ j -1 < i, i ^ j -1 < j, i ^ j -1 < 1, 0 < j, 1 < i, i ^ j -1 + 1 < i, 2 ≤ i, i ^ j -1 ≤ 0, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Proof step 8:
j = i0 +1, i ^ i0 + 1 < i, 1 < i, 0 < j, i ^ i0 < 1, i ^ i0 < j, i ^ i0 < i, i ^ i0 ≤ 0, 2 ≤ i, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Rule argument

spec: int-basic1, name:elim-pred, seq: ⊦ i = j -1 ↔ j = i +1, subst:[j, i] → [j, i0]

Additional information

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

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

j = i0 +1, i ^ i0 + 1 < i, 1 < i, 0 < j, i ^ i0 < 1, i ^ i0 < j, i ^ i0 < i, i ^ i0 ≤ 0, 2 ≤ i, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
i ^ i0 < i, i ^ i0 < 1, 1 < i, i ^ i0 + 1 < i, i ^ i0 ≤ i0, i ^ i0 ≤ 0, 2 ≤ i, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 +1) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i +1 = i + 1; (s) ⊦ ~1 < i ↔ 0 ≤ i; (s) forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (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 numlef-01: ⊦ (* i) ≤ j → i - 1 < 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 ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s) ⊦ i < j + 1 ↔ i ≤ j; (s)

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

i ^ i0 < i, i ^ i0 < 1, 1 < i, i ^ i0 + 1 < i, i ^ i0 ≤ i0, i ^ i0 ≤ 0, 2 ≤ i, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 +1) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

User applied rule apply induction

And closed the goal!

Rule argument

subst: [j0, i] → [i0, i]

Additional information

Used simplifier rules: Pot-zero: ⊦ j ^ 0 = 1; ⊦ i ≤ j ↔ ¬ j < i; (ss) ⊦ i < j → ¬ j < i; (s) ⊦ i < j ↔ ¬ j ≤ i; (ss) ⊦ i ≤ j → (j ≤ i ↔ i = j); (s) ⊦ i +1 = i + 1; (s) ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1; (s) add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 add-c: ⊦ i + j = j + i; from specification int-basic2 ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); (s) ⊦ 0 ≤ i → abs(i) = i; (s) ⊦ ¬ i < i; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: i0, i

Proof Tree for notone (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.