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

⊦ m ≠ 1 → (m * m0 = 3 ↔ m = 3 ∧ m0 = 1)

System applied rule normalize

And obtained the following new goals

Proof step 2:
¬(m * m0 = 3 ↔ m = 3 ∧ m0 = 1), m ≠ 1 ⊦

Rule argument

Additional information

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

¬(m * m0 = 3 ↔ m = 3 ∧ m0 = 1), m ≠ 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m * m0 = 3, m ≠ 1 ⊦ m = 3 ∧ m0 = 1

Rule argument

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n; mult-c: ⊦ m * n = n * m;

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

m * m0 = 3, m ≠ 1 ⊦ m = 3 ∧ m0 = 1

User applied rule cut formula

And obtained the following new goals

Proof step 4:
m * m0 = 3, m ≠ 0, m ≠ 1 ⊦ m = 3 ∧ m0 = 1

Rule argument

fma: m = 0

Additional information

Used simplifier rules: zero: ⊦ 0 * n = 0; mult-a: ⊦ m0 * m * n = (m0 * m) * n; mult-c: ⊦ m * n = n * m; Used formulas: m = 0

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

m * m0 = 3, m ≠ 0, m ≠ 1 ⊦ m = 3 ∧ m0 = 1

User applied rule cut formula

And obtained the following new goals

Proof step 5:
m * m0 = 3, m ≠ 0, m ≠ 1, m0 ≠ 0 ⊦ m = 3 ∧ m0 = 1

Rule argument

fma: m0 = 0

Additional information

Used simplifier rules: zero: ⊦ 0 * n = 0; mult-a: ⊦ m0 * m * n = (m0 * m) * n; mult-c: ⊦ m * n = n * m; Used formulas: m0 = 0

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

m * m0 = 3, m ≠ 0, m ≠ 1, m0 ≠ 0 ⊦ m = 3 ∧ m0 = 1

User applied rule cut formula

And obtained the following new goals

Proof step 6:
m * m0 = 3, m ≠ 0, m ≠ 1, m0 ≠ 1, m0 ≠ 0 ⊦ m = 3 ∧ m0 = 1

Rule argument

fma: m0 = 1

Additional information

Used simplifier rules: one: ⊦ m0 * 1 = m0; mult-a: ⊦ m0 * m * n = (m0 * m) * n; mult-c: ⊦ m * n = n * m; Used formulas: m0 = 1

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

m * m0 = 3, m ≠ 0, m ≠ 1, m0 ≠ 1, m0 ≠ 0 ⊦ m = 3 ∧ m0 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m * m0 = 3, m ≠ 0, m ≠ 1, m0 ≠ 1, m0 ≠ 0 ⊦

Rule argument

Additional information

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

m * m0 = 3, m ≠ 0, m ≠ 1, m0 ≠ 1, m0 ≠ 0 ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 8:
m * m0 = 3, m ≠ 2, m ≠ 0, m ≠ 1, m0 ≠ 1, m0 ≠ 0 ⊦

Rule argument

fma: m = 2

Additional information

Used simplifier rules: threeeq: ⊦ 3 ≠ 2 * n; mult-a: ⊦ m0 * m * n = (m0 * m) * n; mult-c: ⊦ m * n = n * m; Used formulas: m = 2

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

m * m0 = 3, m ≠ 2, m ≠ 0, m ≠ 1, m0 ≠ 1, m0 ≠ 0 ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 9:
m * m0 = 3, m ≠ 2, m ≠ 0, m ≠ 1, m0 ≠ 2, m0 ≠ 1, m0 ≠ 0 ⊦

Rule argument

fma: m0 = 2

Additional information

Used simplifier rules: threeeq: ⊦ 3 ≠ 2 * n; mult-a: ⊦ m0 * m * n = (m0 * m) * n; mult-c: ⊦ m * n = n * m; Used formulas: m0 = 2

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

m * m0 = 3, m ≠ 2, m ≠ 0, m ≠ 1, m0 ≠ 2, m0 ≠ 1, m0 ≠ 0 ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 10:
m * m0 = 3, m ≠ 2, m ≠ 0, m ≠ 1, m0 ≠ 2, m0 ≠ 1, m0 ≠ 0, ¬ 3 < m * m0 ⊦

Rule argument

fma: 3 < m * m0

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic Used formulas: 3 < m * m0

Proof step 10 from proof three (previous step) (back to tree)

m * m0 = 3, m ≠ 2, m ≠ 0, m ≠ 1, m0 ≠ 2, m0 ≠ 1, m0 ≠ 0, ¬ 3 < m * m0 ⊦

User applied rule weakening

And obtained the following new goals

Proof step 11:
m ≠ 2, m ≠ 0, m ≠ 1, m0 ≠ 2, m0 ≠ 1, m0 ≠ 0, ¬ 3 < m * m0 ⊦

Rule argument

poslist: left 1

Additional information

Proof step 11 from proof three (previous step) (back to tree)

m ≠ 2, m ≠ 0, m ≠ 1, m0 ≠ 2, m0 ≠ 1, m0 ≠ 0, ¬ 3 < m * m0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m ≠ 2, m ≠ 0, m ≠ 1, m0 ≠ 2, m0 ≠ 1, m0 ≠ 0, m * m0 ≤ 3 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 12 from proof three (previous step) (back to tree)

m ≠ 2, m ≠ 0, m ≠ 1, m0 ≠ 2, m0 ≠ 1, m0 ≠ 0, m * m0 ≤ 3 ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 13:
m1 +1 ≠ 2, m1 +1 ≠ 1, m1 +1 ≠ 0, m ≠ 2, m ≠ 0, m ≠ 1, m * m1 +1 ≤ 3, ∀ m. ¬(m ≠ 2 ∧ m ≠ 0 ∧ m ≠ 1 ∧ m1 ≠ 2 ∧ m1 ≠ 1 ∧ m1 ≠ 0 ∧ m * m1 ≤ 3) ⊦

Rule argument

indvar: m0

Additional information

Used simplifier rules: nat freely generated by 0, +1;;

Proof step 13 from proof three (previous step) (back to tree)

m1 +1 ≠ 2, m1 +1 ≠ 1, m1 +1 ≠ 0, m ≠ 2, m ≠ 0, m ≠ 1, m * m1 +1 ≤ 3, ∀ m. ¬(m ≠ 2 ∧ m ≠ 0 ∧ m ≠ 1 ∧ m1 ≠ 2 ∧ m1 ≠ 1 ∧ m1 ≠ 0 ∧ m * m1 ≤ 3) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
1 ≠ m1, m ≠ 2, m ≠ 0, m ≠ 1, m1 ≠ 0, m1 * m + m ≤ 3, ∀ m0. m0 = 2 ∨ m0 = 0 ∨ m0 = 1 ∨ m1 = 2 ∨ ¬ m0 * m1 ≤ 3 ⊦

Rule argument

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n; mult-a: ⊦ m0 * m * n = (m0 * m) * n; mult-c: ⊦ m * n = n * m; c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m +1 = m + 1; (s) ⊦ m + 1 ≠ 0; (s) ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; (ws) n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; (s)

Proof step 14 from proof three (previous step) (back to tree)

1 ≠ m1, m ≠ 2, m ≠ 0, m ≠ 1, m1 ≠ 0, m1 * m + m ≤ 3, ∀ m0. m0 = 2 ∨ m0 = 0 ∨ m0 = 1 ∨ m1 = 2 ∨ ¬ m0 * m1 ≤ 3 ⊦

User applied rule constructor cut

And obtained the following new goals

Proof step 15:
m = m2 +1, 1 ≠ m1, m ≠ 2, m ≠ 0, m ≠ 1, m1 ≠ 0, m1 * m + m ≤ 3, ∀ m0. m0 = 2 ∨ m0 = 0 ∨ m0 = 1 ∨ m1 = 2 ∨ ¬ m0 * m1 ≤ 3 ⊦

Rule argument

var: m

Additional information

Used simplifier rules: nat freely generated by 0, +1;;

Proof step 15 from proof three (previous step) (back to tree)

m = m2 +1, 1 ≠ m1, m ≠ 2, m ≠ 0, m ≠ 1, m1 ≠ 0, m1 * m + m ≤ 3, ∀ m0. m0 = 2 ∨ m0 = 0 ∨ m0 = 1 ∨ m1 = 2 ∨ ¬ m0 * m1 ≤ 3 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
1 ≠ m1, 1 ≠ m2, m1 ≠ 0, m2 ≠ 0, m2 * m1 + m1 + m2 ≤ 2, ∀ m0. m0 = 2 ∨ m0 = 0 ∨ m0 = 1 ∨ m1 = 2 ∨ ¬ m0 * m1 ≤ 3 ⊦

Rule argument

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n; mult-a: ⊦ m0 * m * n = (m0 * m) * n; mult-c: ⊦ m * n = n * m; c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic n < n0 ⊦ m + (* n) ≤ (* n0) ↔ m ≤ n0 - n; (s) ⊦ m +1 = m + 1; (s) ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; (ws) ⊦ m + 1 ≠ 0; (s) n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; (s)

Proof step 16 from proof three (previous step) (back to tree)

1 ≠ m1, 1 ≠ m2, m1 ≠ 0, m2 ≠ 0, m2 * m1 + m1 + m2 ≤ 2, ∀ m0. m0 = 2 ∨ m0 = 0 ∨ m0 = 1 ∨ m1 = 2 ∨ ¬ m0 * m1 ≤ 3 ⊦

User applied rule constructor cut

And closed the goal!

Rule argument

var: m1

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n; two-02: ⊦ m + m = 2 * m; one: ⊦ m0 * 1 = m0; succ: ⊦ (* m0) * m + m = (m0 + 1) * m; zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); mult-a: ⊦ m0 * m * n = (m0 * m) * n; mult-c: ⊦ m * n = n * m; nat freely generated by 0, +1;; ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; (ws) ⊦ m + 1 ≠ 0; (s) ⊦ m +1 = m + 1; (s) n < n0 ⊦ m + (* n) ≤ (* n0) ↔ m ≤ n0 - n; (s) ⊦ m ≠ 0 → (m + n ≤ 1 ↔ m = 1 ∧ n = 0); (ws) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

Proof Tree for three (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.