Proof step 1 from proof geom-mod (conclusion) (back to tree)

⊦ i ≠ 1 ∧ 0 ≤ j → (1 - i ^ j) % (1 - i) = 0

System applied rule normalize

And obtained the following new goals

Proof step 2:
(1 - i ^ j) % (1 - i) ≠ 0, i ≠ 1, 0 ≤ j ⊦

Rule argument

Additional information

Proof step 2 from proof geom-mod (previous step) (back to tree)

(1 - i ^ j) % (1 - i) ≠ 0, i ≠ 1, 0 ≤ j ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
(1 - i ^ j) % (1 - i) ≠ 0, i ≠ 1, 0 ≤ j ⊦

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs

Proof step 3 from proof geom-mod (previous step) (back to tree)

(1 - i ^ j) % (1 - i) ≠ 0, i ≠ 1, 0 ≤ j ⊦

User applied rule induction

And obtained the following new goals

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

Rule argument

indvar: i→n(j) wfpred: <

Additional information

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

Proof step 4 from proof geom-mod (previous step) (back to tree)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
(1 - i ^ j) % (1 - i) ≠ 0, i ≠ 1, 0 ≤ j, ∀ j0, i. i→n(j0) < i→n(j) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

Additional information

Proof step 5 from proof geom-mod (previous step) (back to tree)

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

User applied rule cut formula

And obtained the following new goals

Proof step 6:
(1 - i ^ j) % (1 - i) ≠ 0, i ≠ 1, j ≠ 0, 0 ≤ j, ∀ j0, i. i→n(j0) < i→n(j) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

fma: j = 0

Additional information

Used simplifier rules: Pot-zero: ⊦ j ^ 0 = 1; ⊦ i - j = i0 ↔ i = i0 + j; (s) ⊦ i + 0 = i; (s) add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 add-c: ⊦ i + j = j + i; from specification int-basic2 j ≠ 0 ⊦ 0 % j = 0; (s) Used formulas: j = 0

Proof step 6 from proof geom-mod (previous step) (back to tree)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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

Proof step 7 from proof geom-mod (previous step) (back to tree)

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

User applied rule apply induction

And obtained the following new goals

Proof step 8:
(1 - i ^ j) % (1 - i) ≠ 0, i ≠ 1, j ≠ 0, 0 < j, ∀ j0, i. i→n(j0) < i→n(j) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦ (1 - i ^ (j - 1)) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j - 1

Rule argument

subst: [j0, i] → [j - 1, i]

Additional information

Used simplifier rules: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; (s) ⊦ i < j ↔ ¬ j ≤ i; (ss) ⊦ i ≠ j → (i ≤ j ↔ i < j); (s) ⊦ i < j → ¬ j < i; (s) Used terms: j - 1, i

Proof step 8 from proof geom-mod (previous step) (back to tree)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
(1 - i ^ j) % (1 - i) ≠ 0, i ≠ 1, j ≠ 0, 0 < j, ∀ j0, i. i→n(j0) < i→n(j) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦ (1 - i ^ (j - 1)) % (1 - i) ≠ 0 ∧ 0 ≤ j - 1

Rule argument

Additional information

Proof step 9 from proof geom-mod (previous step) (back to tree)

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

User applied rule apply elim lemma

And obtained the following new goals

Proof step 10:
j = i0 + 1, (1 - i ^ j) % (1 - i) ≠ 0, i ≠ 1, j ≠ 0, 0 < j, ∀ j0, i. i→n(j0) < i→n(j) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦ (1 - i ^ i0) % (1 - i) ≠ 0 ∧ 0 ≤ i0

Rule argument

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

Additional information

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

Proof step 10 from proof geom-mod (previous step) (back to tree)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(1 - i ^ i0) % (1 - i) = 0, (1 - i ^ i0 * i) % (1 - i) ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

Additional information

Used simplifier rules: succ: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j; 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 ⊦ ~1 < i ↔ 0 ≤ i; (s) ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s) ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)

Proof step 11 from proof geom-mod (previous step) (back to tree)

(1 - i ^ i0) % (1 - i) = 0, (1 - i ^ i0 * i) % (1 - i) ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 12:
(1 - i ^ i0) % (1 - i) = 0, 1 - i ^ i0 * i = (1 - i ^ i0) * i - (i - 1), (1 - i ^ i0 * i) % (1 - i) ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

fma: 1 - i ^ i0 * i = (1 - i ^ i0) * i - (i - 1)

Additional information

Used simplifier rules: ⊦ i - (j - j0) = i + j0 - j; (s) ⊦ i - j = i0 ↔ i = i0 + j; (s) ⊦ i + (j - i0) = i + j - i0; (s) ⊦ i * j + i * j0 = i * (j + j0); (s) ⊦ i + (j - i) = j; (s) ⊦ 1 * i = i; (s) associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult commutativity: ⊦ i * j = j * i; from specification int-mult ⊦ 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 Used formulas: 1 - i ^ i0 * i = (1 - i ^ i0) * i - (i - 1)

Proof step 12 from proof geom-mod (previous step) (back to tree)

(1 - i ^ i0) % (1 - i) = 0, 1 - i ^ i0 * i = (1 - i ^ i0) * i - (i - 1), (1 - i ^ i0 * i) % (1 - i) ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule subst equation

And obtained the following new goals

Proof step 13:
(1 - i ^ i0) % (1 - i) = 0, ((1 - i ^ i0) * i - (i - 1)) % (1 - i) ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

pos: left 2, flags: discard

Additional information

Proof step 13 from proof geom-mod (previous step) (back to tree)

(1 - i ^ i0) % (1 - i) = 0, ((1 - i ^ i0) * i - (i - 1)) % (1 - i) ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 14:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)
Proof step 15:
((1 - i ^ i0) * i - (i - 1)) % (1 - i) ≠ 0, 1 - i ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∃ j0. 1 - i ^ i0 = j0 * (1 - i), ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

spec: int-div, name:modex, seq:j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j), subst:[j, i] → [1 - i, 1 - i ^ i0], <all paths>

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j; (s) ⊦ i + 0 = i; (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 → i - j = 0; (s) Used terms: 1 - i, 1 - i ^ i0

Proof step 14 from proof geom-mod (previous step) (back to tree)

j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)

A spec-lemma

Proof step 15 from proof geom-mod (previous step) (back to tree)

((1 - i ^ i0) * i - (i - 1)) % (1 - i) ≠ 0, 1 - i ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∃ j0. 1 - i ^ i0 = j0 * (1 - i), ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule exists left

And obtained the following new goals

Proof step 16:
1 - i ^ i0 = j0 * (1 - i), ((1 - i ^ i0) * i - (i - 1)) % (1 - i) ≠ 0, 1 - i ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

pos: left 6

Additional information

Proof step 16 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = j0 * (1 - i), ((1 - i ^ i0) * i - (i - 1)) % (1 - i) ≠ 0, 1 - i ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule subst equation

And obtained the following new goals

Proof step 17:
1 - i ^ i0 = j0 * (1 - i), ((j0 * (1 - i)) * i - (i - 1)) % (1 - i) ≠ 0, 1 - i ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

pos: left 1, flags:

Additional information

Proof step 17 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = j0 * (1 - i), ((j0 * (1 - i)) * i - (i - 1)) % (1 - i) ≠ 0, 1 - i ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 18:
j ≠ 0 ⊦ i % ~ j = i % j
Proof step 19:
1 - i ^ i0 = j0 * (1 - i), ((j0 * (1 - i)) * i - (i - 1)) % ~(1 - i) ≠ 0, 1 - i ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

spec: int-div, name:mod-neg, seq:j ≠ 0 ⊦ i % ~ j = i % j, subst:[j, i] → [1 - i, (j0 * (1 - i)) * i - (i - 1)], <rotated> <all paths>

Additional information

Used terms: 1 - i, (j0 * (1 - i)) * i - (i - 1)

Proof step 18 from proof geom-mod (previous step) (back to tree)

j ≠ 0 ⊦ i % ~ j = i % j

A spec-lemma

Proof step 19 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = j0 * (1 - i), ((j0 * (1 - i)) * i - (i - 1)) % ~(1 - i) ≠ 0, 1 - i ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ ~(i - j) = j - i
Proof step 21:
1 - i ^ i0 = j0 * (1 - i), ((j0 * (1 - i)) * i - (i - 1)) % (i - 1) ≠ 0, 1 - i ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

spec: int-abs, name:minus-sub, seq: ⊦ ~(i - j) = j - i, subst:[j, i] → [i, 1], <all paths>

Additional information

Used terms: i, 1

Proof step 20 from proof geom-mod (previous step) (back to tree)

⊦ ~(i - j) = j - i

A spec-lemma

Proof step 21 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = j0 * (1 - i), ((j0 * (1 - i)) * i - (i - 1)) % (i - 1) ≠ 0, 1 - i ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 22:
1 - i ^ i0 = j0 * (1 - i), 1 - i = ~ ~(1 - i), ((j0 * (1 - i)) * i - (i - 1)) % (i - 1) ≠ 0, 1 - i ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

fma: 1 - i = ~ ~(1 - i)

Additional information

Used simplifier rules: ⊦ ~(i - j) = j - i; (s) ⊦ ~(i - 1) = ~ i + 1; (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 Used formulas: 1 - i = ~ ~(1 - i)

Proof step 22 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = j0 * (1 - i), 1 - i = ~ ~(1 - i), ((j0 * (1 - i)) * i - (i - 1)) % (i - 1) ≠ 0, 1 - i ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule subst equation

And obtained the following new goals

Proof step 23:
1 - i ^ i0 = j0 * ~ ~(1 - i), ((j0 * ~ ~(1 - i)) * i - (i - 1)) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~ ~(1 - i) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

pos: left 2, flags: discard

Additional information

Proof step 23 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = j0 * ~ ~(1 - i), ((j0 * ~ ~(1 - i)) * i - (i - 1)) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~ ~(1 - i) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 24:
⊦ ~(i - j) = j - i
Proof step 25:
1 - i ^ i0 = j0 * ~(i - 1), ((j0 * ~(i - 1)) * i - (i - 1)) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~(i - 1) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

spec: int-abs, name:minus-sub, seq: ⊦ ~(i - j) = j - i, subst:[j, i] → [i, 1], <all paths>

Additional information

Used terms: i, 1

Proof step 24 from proof geom-mod (previous step) (back to tree)

⊦ ~(i - j) = j - i

A spec-lemma

Proof step 25 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = j0 * ~(i - 1), ((j0 * ~(i - 1)) * i - (i - 1)) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~(i - 1) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 26:
⊦ ~ i * j = ~(i * j)
Proof step 27:
1 - i ^ i0 = ~((i - 1) * j0), (~((i - 1) * j0) * i - (i - 1)) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~(i - 1) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

spec: int-mult, name:minus-left, seq: ⊦ ~ i * j = ~(i * j), subst:[j, i] → [j0, i - 1], <all paths>

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j0, i - 1

Proof step 26 from proof geom-mod (previous step) (back to tree)

⊦ ~ i * j = ~(i * j)

A spec-lemma

Proof step 27 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = ~((i - 1) * j0), (~((i - 1) * j0) * i - (i - 1)) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~(i - 1) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 28:
⊦ ~ i - ~ j = j - i
Proof step 29:
1 - i ^ i0 = ~((i - 1) * j0), (~(i - 1) - ~(~((i - 1) * j0) * i)) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~(i - 1) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

spec: int-abs, name:sub-minus, seq: ⊦ ~ i - ~ j = j - i, subst:[i, j] → [i - 1, ~((i - 1) * j0) * i], <rotated> <all paths>

Additional information

Used terms: i - 1, ~((i - 1) * j0) * i

Proof step 28 from proof geom-mod (previous step) (back to tree)

⊦ ~ i - ~ j = j - i

A spec-lemma

Proof step 29 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = ~((i - 1) * j0), (~(i - 1) - ~(~((i - 1) * j0) * i)) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~(i - 1) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ ~ i * j = ~(i * j)
Proof step 31:
1 - i ^ i0 = ~((i - 1) * j0), (~(i - 1) - ~ ~(((i - 1) * j0) * i)) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~(i - 1) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

spec: int-mult, name:minus-left, seq: ⊦ ~ i * j = ~(i * j), subst:[j, i] → [i, (i - 1) * j0], <all paths>

Additional information

Used terms: i, (i - 1) * j0

Proof step 30 from proof geom-mod (previous step) (back to tree)

⊦ ~ i * j = ~(i * j)

A spec-lemma

Proof step 31 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = ~((i - 1) * j0), (~(i - 1) - ~ ~(((i - 1) * j0) * i)) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~(i - 1) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 32:
⊦ ~ ~ i = i
Proof step 33:
1 - i ^ i0 = ~((i - 1) * j0), (~(i - 1) - ((i - 1) * j0) * i) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~(i - 1) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

spec: int-abs, name:neg-02, seq: ⊦ ~ ~ i = i, subst:[i] → [((i - 1) * j0) * i], <all paths>

Additional information

Used terms: ((i - 1) * j0) * i

Proof step 32 from proof geom-mod (previous step) (back to tree)

⊦ ~ ~ i = i

A spec-lemma

Proof step 33 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = ~((i - 1) * j0), (~(i - 1) - ((i - 1) * j0) * i) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~(i - 1) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 34:
1 - i ^ i0 = ~((i - 1) * j0), ~(i - 1) = (i - 1) * ~1, (~(i - 1) - ((i - 1) * j0) * i) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~(i - 1) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

fma: ~(i - 1) = (i - 1) * ~1

Additional information

Used simplifier rules: ⊦ i * ~1 = ~ i; (s) ⊦ ~(i - 1) = ~ i + 1; (s) ⊦ i + ~ j = i - j; (s) associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult commutativity: ⊦ i * j = j * i; from specification int-mult ⊦ 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 Used formulas: ~(i - 1) = (i - 1) * ~1

Proof step 34 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = ~((i - 1) * j0), ~(i - 1) = (i - 1) * ~1, (~(i - 1) - ((i - 1) * j0) * i) % (i - 1) ≠ 0, ~1 ≠ i0, i ≠ 1, ~(i - 1) ≠ 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule subst equation

And obtained the following new goals

Proof step 35:
1 - i ^ i0 = ~((i - 1) * j0), ((i - 1) * ~1 - ((i - 1) * j0) * i) % (i - 1) ≠ 0, (i - 1) * ~1 ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

Rule argument

pos: left 2, flags: discard

Additional information

Proof step 35 from proof geom-mod (previous step) (back to tree)

1 - i ^ i0 = ~((i - 1) * j0), ((i - 1) * ~1 - ((i - 1) * j0) * i) % (i - 1) ≠ 0, (i - 1) * ~1 ≠ 0, ~1 ≠ i0, i ≠ 1, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → (1 - i ^ j0) % (1 - i) ≠ 0 ∧ i ≠ 1 ∧ 0 ≤ j0 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 36:
⊦ i * j - i * j0 = i * (j - j0)

Rule argument

spec: int-mult, name:dist-sub-merge, seq: ⊦ i * j - i * j0 = i * (j - j0), subst:[j0, j, i] → [i * j0, ~1, i - 1], <all paths>

Additional information

Used simplifier rules: ⊦ i + ~ j = i - j; (s) ⊦ i - j = i0 ↔ i = i0 + j; (s) add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 add-c: ⊦ i + j = j + i; from specification int-basic2 j ≠ 0 ⊦ (i * j) % 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: i * j0, ~1, i - 1

Proof step 36 from proof geom-mod (previous step) (back to tree)

⊦ i * j - i * j0 = i * (j - j0)

A spec-lemma

Proof Tree for geom-mod (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.