[Imprint][Data Privacy Statement][E-Mail]



View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 1 from proof mod-by6 (conclusion) (back to tree)

⊦ (n * (n + 1) * (2 * n + 1)) % 6 = 0

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof mod-by6 (previous step) (back to tree)

(n * (n + 1) * (2 * n + 1)) % 6 ≠ 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m * (n + 1) = m * n + m; (s)

Proof step 3 from proof mod-by6 (previous step) (back to tree)

((n * n + n) * 2 * n + n * n + n) % 6 ≠ 0 ⊦

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Rule argument

term: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;; mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic

Proof step 4 from proof mod-by6 (previous step) (back to tree)

((n0 * n0 + n0) * 2 * n0 + n0 * n0 + n0) % 6 = 0, ((n0 +1 * n0 +1 + n0 +1) * 2 * n0 +1 + n0 +1 * n0 +1 + n0 +1) % 6 ≠ 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult ⊦ (* m0) * m + m = (m0 + 1) * m; (s) ⊦ m +1 = m + 1; (s) ⊦ m + m = 2 * m; (s) ⊦ m * (n + 1) = m * n + m; (s)

Proof step 5 from proof mod-by6 (previous step) (back to tree)

((n0 * n0 + n0) * 2 * n0 + n0 * n0 + n0) % 6 = 0, (3 * n0 + 2 + ((3 * n0 + 2 + n0 * n0) * n0 + 3 * n0 + 2 + n0 * n0) * 2 + n0 * n0) % 6 ≠ 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-mult, name:dist, seq: ⊦ m0 * (m + n) = m0 * m + m0 * n, subst:[m0, n, m] → [n0 * 2, n0, n0 * n0], <all paths>

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: n0 * 2, n0, n0 * n0

Proof step 6 from proof mod-by6 (previous step) (back to tree)

⊦ m0 * (m + n) = m0 * m + m0 * n

A spec-lemma

Proof step 7 from proof mod-by6 (previous step) (back to tree)

(((n0 * 2) * n0 * n0 + (n0 * 2) * n0) + n0 * n0 + n0) % 6 = 0, (3 * n0 + 2 + ((3 * n0 + 2 + n0 * n0) * n0 + 3 * n0 + 2 + n0 * n0) * 2 + n0 * n0) % 6 ≠ 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult ⊦ (* m0) * m + m = (m0 + 1) * m; (s)

Proof step 8 from proof mod-by6 (previous step) (back to tree)

(3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) % 6 = 0, (3 * n0 + 2 + ((3 * n0 + 2 + n0 * n0) * n0 + 3 * n0 + 2 + n0 * n0) * 2 + n0 * n0) % 6 ≠ 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-mult, name:dist, seq: ⊦ m0 * (m + n) = m0 * m + m0 * n, subst:[m0, n, m] → [n0, 2 + n0 * n0, 3 * n0], <all paths>

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: n0, 2 + n0 * n0, 3 * n0

Proof step 9 from proof mod-by6 (previous step) (back to tree)

⊦ m0 * (m + n) = m0 * m + m0 * n

A spec-lemma

Proof step 10 from proof mod-by6 (previous step) (back to tree)

(3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) % 6 = 0, (3 * n0 + 2 + ((n0 * 3 * n0 + n0 * (2 + n0 * n0)) + 3 * n0 + 2 + n0 * n0) * 2 + n0 * n0) % 6 ≠ 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult ⊦ (* m0) * m + m = (m0 + 1) * m; (s)

Proof step 11 from proof mod-by6 (previous step) (back to tree)

(3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) % 6 = 0, (3 * n0 + 2 + (4 * n0 * n0 + n0 * (2 + n0 * n0) + 3 * n0 + 2) * 2 + n0 * n0) % 6 ≠ 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-mult, name:dist, seq: ⊦ m0 * (m + n) = m0 * m + m0 * n, subst:[n, m, m0] → [n0 * n0, 2, n0], <all paths>

Additional information

Used terms: n0 * n0, 2, n0

Proof step 12 from proof mod-by6 (previous step) (back to tree)

⊦ m0 * (m + n) = m0 * m + m0 * n

A spec-lemma

Proof step 13 from proof mod-by6 (previous step) (back to tree)

(3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) % 6 = 0, (3 * n0 + 2 + (4 * n0 * n0 + (n0 * 2 + n0 * n0 * n0) + 3 * n0 + 2) * 2 + n0 * n0) % 6 ≠ 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult ⊦ m0 * (* m) + m0 * (* n) = m0 * (m + n); (s)

Proof step 14 from proof mod-by6 (previous step) (back to tree)

(3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) % 6 = 0, (3 * n0 + 2 + (n0 * 5 + 4 * n0 * n0 + n0 * n0 * n0 + 2) * 2 + n0 * n0) % 6 ≠ 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-mult, name:dist, seq: ⊦ m0 * (m + n) = m0 * m + m0 * n, subst:[m0, n, m] → [2, 4 * n0 * n0 + n0 * n0 * n0 + 2, n0 * 5], <all paths>

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: 2, 4 * n0 * n0 + n0 * n0 * n0 + 2, n0 * 5

Proof step 15 from proof mod-by6 (previous step) (back to tree)

⊦ m0 * (m + n) = m0 * m + m0 * n

A spec-lemma

Proof step 16 from proof mod-by6 (previous step) (back to tree)

(3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) % 6 = 0, (3 * n0 + 2 + (2 * n0 * 5 + 2 * (4 * n0 * n0 + n0 * n0 * n0 + 2)) + n0 * n0) % 6 ≠ 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult ⊦ m0 * (* m) + m0 * (* n) = m0 * (m + n); (s)

Proof step 17 from proof mod-by6 (previous step) (back to tree)

(3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) % 6 = 0, (n0 * 13 + 2 + 2 * (4 * n0 * n0 + n0 * n0 * n0 + 2) + n0 * n0) % 6 ≠ 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-mult, name:dist, seq: ⊦ m0 * (m + n) = m0 * m + m0 * n, subst:[n, m, m0] → [n0 * n0 * n0 + 2, 4 * n0 * n0, 2], <all paths>

Additional information

Used terms: n0 * n0 * n0 + 2, 4 * n0 * n0, 2

Proof step 18 from proof mod-by6 (previous step) (back to tree)

⊦ m0 * (m + n) = m0 * m + m0 * n

A spec-lemma

Proof step 19 from proof mod-by6 (previous step) (back to tree)

(3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) % 6 = 0, (n0 * 13 + 2 + (2 * 4 * n0 * n0 + 2 * (n0 * n0 * n0 + 2)) + n0 * n0) % 6 ≠ 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult ⊦ (* m0) * m + m = (m0 + 1) * m; (s)

Proof step 20 from proof mod-by6 (previous step) (back to tree)

(3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) % 6 = 0, (9 * n0 * n0 + n0 * 13 + 2 + 2 * (n0 * n0 * n0 + 2)) % 6 ≠ 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-mult, name:dist, seq: ⊦ m0 * (m + n) = m0 * m + m0 * n, subst:[n, m, m0] → [2, n0 * n0 * n0, 2], <all paths>

Additional information

Used terms: 2, n0 * n0 * n0, 2

Proof step 21 from proof mod-by6 (previous step) (back to tree)

⊦ m0 * (m + n) = m0 * m + m0 * n

A spec-lemma

Proof step 22 from proof mod-by6 (previous step) (back to tree)

(3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) % 6 = 0, (9 * n0 * n0 + n0 * 13 + 2 + 2 * n0 * n0 * n0 + 2 * 2) % 6 ≠ 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mod-once: n ≠ 0 ⊦ (m + n) % n = m % n; c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult

Proof step 23 from proof mod-by6 (previous step) (back to tree)

(3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) % 6 = 0, (9 * n0 * n0 + n0 * 13 + 2 * n0 * n0 * n0) % 6 ≠ 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:modex, seq: ⊦ n ≠ 0 → (m % n = 0 ↔ (∃ n0. m = n0 * n)), subst:[n, m] → [6, 3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0], <all paths>

Additional information

Used terms: 6, 3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0

Proof step 24 from proof mod-by6 (previous step) (back to tree)

⊦ n ≠ 0 → (m % n = 0 ↔ (∃ n0. m = n0 * n))

Lemma modex

Proof step 25 from proof mod-by6 (previous step) (back to tree)

(9 * n0 * n0 + n0 * 13 + 2 * n0 * n0 * n0) % 6 ≠ 0, 6 ≠ 0, ∃ n1. 3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0 = n1 * 6 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 26 from proof mod-by6 (previous step) (back to tree)

3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0 = n1 * 6, (9 * n0 * n0 + n0 * 13 + 2 * n0 * n0 * n0) % 6 ≠ 0 ⊦

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: 9 * n0 * n0 + n0 * 13 + 2 * n0 * n0 * n0 = (3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) + 6 * (n0 * n0 + 2 * n0)

Additional information

Used formulas: 9 * n0 * n0 + n0 * 13 + 2 * n0 * n0 * n0 = (3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) + 6 * (n0 * n0 + 2 * n0)

Proof step 27 from proof mod-by6 (previous step) (back to tree)

3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0 = n1 * 6, 9 * n0 * n0 + n0 * 13 + 2 * n0 * n0 * n0 = (3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) + 6 * (n0 * n0 + 2 * n0), (9 * n0 * n0 + n0 * 13 + 2 * n0 * n0 * n0) % 6 ≠ 0 ⊦

User applied rule subst equation

And obtained the following new goals

Rule argument

pos: left 2, flags: discard

Additional information

Proof step 28 from proof mod-by6 (previous step) (back to tree)

3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0 = n1 * 6, ((3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) + 6 * (n0 * n0 + 2 * n0)) % 6 ≠ 0 ⊦

User applied rule subst equation

And obtained the following new goals

Rule argument

pos: left 1, flags: discard

Additional information

Proof step 29 from proof mod-by6 (previous step) (back to tree)

(n1 * 6 + 6 * (n0 * n0 + 2 * n0)) % 6 ≠ 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:mod-add, seq: ⊦ m0 % n = 0 ∧ 0 ≠ n → (m + m0) % n = m % n, subst:[n, m0, m] → [6, 6 * (n0 * n0 + 2 * n0), n1 * 6], <all paths>

Additional information

Used simplifier rules: mod-mult-01: n ≠ 0 ⊦ (m * n) % n = 0; mod-mult-01: n ≠ 0 ⊦ (m * n) % n = 0; mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: 6, 6 * (n0 * n0 + 2 * n0), n1 * 6

Proof step 30 from proof mod-by6 (previous step) (back to tree)

⊦ m0 % n = 0 ∧ 0 ≠ n → (m + m0) % n = m % n

Lemma mod-add

Proof step 31 from proof mod-by6 (previous step) (back to tree)

3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0 = n1 * 6, (9 * n0 * n0 + n0 * 13 + 2 * n0 * n0 * n0) % 6 ≠ 0, 9 * n0 * n0 + n0 * 13 + 2 * n0 * n0 * n0 ≠ (3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0) + 6 * (n0 * n0 + 2 * n0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult m2 ≠ 0 ⊦ (* m2) * m + n = m + n0 ↔ (m2 - 1) * m + n = n0; (s) m1 < m2 ⊦ (* m2) * m + n = (* m1) * m + n0 ↔ (m2 - m1) * m + n = n0; (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 32 from proof mod-by6 (previous step) (back to tree)

3 * n0 * n0 + (n0 * 2) * n0 * n0 + n0 = n1 * 6, (9 * n0 * n0 + n0 * 13 + 2 * n0 * n0 * n0) % 6 ≠ 0, 12 * n0 + 6 * n0 * n0 ≠ 6 * (n0 * n0 + 2 * n0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-mult, name:dist, seq: ⊦ m0 * (m + n) = m0 * m + m0 * n, subst:[n, m, m0] → [2 * n0, n0 * n0, 6], <all paths>

Additional information

Used simplifier rules: ⊦ m * n + m0 * n = m1 * n + m2 * n ↔ m + m0 = m1 + m2 ∨ n = 0; (s) mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: 2 * n0, n0 * n0, 6

Proof step 33 from proof mod-by6 (previous step) (back to tree)

⊦ m0 * (m + n) = m0 * m + m0 * n

A spec-lemma

Proof Tree for mod-by6 (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