Proof step 1 from proof mod2-2pot (conclusion) (back to tree)

⊦ 0 < i → (j % 2 ^ i) % 2 = j % 2

System applied rule normalize

And obtained the following new goals

Proof step 2:
(j % 2 ^ i) % 2 ≠ j % 2, 0 < i ⊦

Rule argument

Additional information

Proof step 2 from proof mod2-2pot (previous step) (back to tree)

(j % 2 ^ i) % 2 ≠ j % 2, 0 < i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
(j % 2 ^ i) % 2 ≠ j % 2, 0 < i ⊦

Rule argument

Additional information

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

Proof step 3 from proof mod2-2pot (previous step) (back to tree)

(j % 2 ^ i) % 2 ≠ j % 2, 0 < i ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < i → j ^ i = j ^ i -1 * j
Proof step 5:
(j % (2 ^ i -1 * 2)) % 2 ≠ j % 2, 0 < i ⊦

Rule argument

spec: <toplevel>, name:Pot-pos, seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j, subst:[i, j] → [i, 2], <all paths>

Additional information

Used terms: i, 2

Proof step 4 from proof mod2-2pot (previous step) (back to tree)

⊦ 0 < i → j ^ i = j ^ i -1 * j

Lemma Pot-pos

Proof step 5 from proof mod2-2pot (previous step) (back to tree)

(j % (2 ^ i -1 * 2)) % 2 ≠ j % 2, 0 < i ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Proof step 6:
i = i0 +1, (j % (2 ^ i0 * 2)) % 2 ≠ j % 2, 0 < i ⊦

Rule argument

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

Additional information

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

Proof step 6 from proof mod2-2pot (previous step) (back to tree)

i = i0 +1, (j % (2 ^ i0 * 2)) % 2 ≠ j % 2, 0 < i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
(j % (2 ^ i0 * 2)) % 2 ≠ j % 2, 0 ≤ i0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ~1 < i ↔ 0 ≤ i; (s) 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 +1 = i + 1; (s)

Proof step 7 from proof mod2-2pot (previous step) (back to tree)

(j % (2 ^ i0 * 2)) % 2 ≠ j % 2, 0 ≤ i0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ i * j = j * i
Proof step 9:
(j % (2 * 2 ^ i0)) % 2 ≠ j % 2, 0 ≤ i0 ⊦

Rule argument

spec: int-mult, name:commutativity, seq: ⊦ i * j = j * i, subst:[j, i] → [2, 2 ^ i0], <all paths>

Additional information

Used terms: 2, 2 ^ i0

Proof step 8 from proof mod2-2pot (previous step) (back to tree)

⊦ i * j = j * i

A spec-lemma

Proof step 9 from proof mod2-2pot (previous step) (back to tree)

(j % (2 * 2 ^ i0)) % 2 ≠ j % 2, 0 ≤ i0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 10:
j ≠ 0 ⊦ 0 < j0 → (i % (j * j0)) % j = i % j

Rule argument

spec: int-div, name:modmod-mult-03, seq:j ≠ 0 ⊦ 0 < j0 → (i % (j * j0)) % j = i % j, subst:[j0, j, i] → [2 ^ i0, 2, j], <all paths>

Additional information

Used simplifier rules: two-lez: ⊦ 0 ≤ i → ¬ 2 ^ i ≤ 0; ⊦ i < j ↔ ¬ j ≤ i; (ss) Used terms: 2 ^ i0, 2, j

Proof step 10 from proof mod2-2pot (previous step) (back to tree)

j ≠ 0 ⊦ 0 < j0 → (i % (j * j0)) % j = i % j

A spec-lemma

Proof Tree for mod2-2pot (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.