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Proof step 1 from proof odd-mult (conclusion) (back to tree)

⊦ odd(m * n) ↔ odd(m) ∧ odd(n)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof odd-mult (previous step) (back to tree)

¬(odd(m * n) ↔ odd(m) ∧ odd(n)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: Odd: ⊦ odd(n) ↔ ¬ even(n); Even: ⊦ even(n) ↔ (∃ m. m + m = n); c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + m = 2 * m; (s)

Proof step 3 from proof odd-mult (previous step) (back to tree)

⊦ ¬(∃ m0. 2 * m0 = m * n) ↔ ¬(∃ m0. 2 * m0 = m) ∧ ¬(∃ m0. 2 * m0 = n)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Proof step 4 from proof odd-mult (previous step) (back to tree)

⊦ ∃ m0. 2 * m0 = m * n, ¬(∃ m0. 2 * m0 = m) ∧ ¬(∃ m0. 2 * m0 = n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 5 from proof odd-mult (previous step) (back to tree)

⊦ ∃ m0. 2 * m0 = m * n, 2 * m0 ≠ m ∧ 2 * m1 ≠ n

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

Additional information

Proof step 6 from proof odd-mult (previous step) (back to tree)

2 * m0 = m ⊦ ∃ m0. 2 * m0 = m * n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 7 from proof odd-mult (previous step) (back to tree)

⊦ ∃ m1. 2 * m1 = (2 * m0) * n

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: n * m0

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: n * m0

Proof step 8 from proof odd-mult (previous step) (back to tree)

2 * m1 = n ⊦ ∃ m0. 2 * m0 = m * n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 9 from proof odd-mult (previous step) (back to tree)

⊦ ∃ m0. 2 * m0 = m * 2 * m1

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: m1 * m

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: m1 * m

Proof step 10 from proof odd-mult (previous step) (back to tree)

∃ m0. 2 * m0 = m * n ⊦ ∃ m0. 2 * m0 = n, ∃ m0. 2 * m0 = m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 11 from proof odd-mult (previous step) (back to tree)

2 * m0 = m * n ⊦ ∃ m0. 2 * m0 = m, ∃ m0. 2 * m0 = n

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 2 terms: n / 2 abort: F discard: T computed: F

Additional information

Used terms: n / 2

Proof step 12 from proof odd-mult (previous step) (back to tree)

2 * m0 = m * n, 2 * n / 2 ≠ n ⊦ ∃ m0. 2 * m0 = m

User applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:div2-elim, seq: ⊦ odd(n) → (m = n / 2 ↔ n = m + m + 1), subst:[n, m] → [n, m1]

Additional information

Used terms: n, m1

Proof step 13 from proof odd-mult (previous step) (back to tree)

⊦ odd(n) → (m = n / 2 ↔ n = m + m + 1)

Lemma div2-elim

Proof step 14 from proof odd-mult (previous step) (back to tree)

2 * m0 = m * n, 2 * n / 2 ≠ n, ¬ odd(n) ⊦ ∃ m0. 2 * m0 = m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: pot2-even-02: ⊦ n / 2 * 2 = n ↔ even(n); Odd: ⊦ odd(n) ↔ ¬ even(n); Even: ⊦ even(n) ↔ (∃ m. m + m = n); c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + m = 2 * m; (s) mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult

Proof step 15 from proof odd-mult (previous step) (back to tree)

2 * m0 = m * 2 * m1 ⊦ ∃ m. 2 * m = 2 * m1, ∃ m0. 2 * m0 = m

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: m1

Additional information

Used terms: m1

Proof step 16 from proof odd-mult (previous step) (back to tree)

2 * m0 = m * n, n = m1 + m1 + 1, 2 * m1 ≠ n, odd(n) ⊦ ∃ m0. 2 * m0 = m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: modd-02: ⊦ odd(2 * m + 1); 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 ⊦ m + m = 2 * m; (s) ⊦ 2 * m ≠ 2 * n + 1; (s) ⊦ m * (n + 1) = m * n + m; (s)

Proof step 17 from proof odd-mult (previous step) (back to tree)

2 * m0 = m * 2 * m1 + m ⊦ ∃ m0. 2 * m0 = m

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: m0 - m * m1 abort: F discard: T computed: F

Additional information

Used terms: m0 - m * m1

Proof step 18 from proof odd-mult (previous step) (back to tree)

2 * m0 = m * 2 * m1 + m, 2 * (m0 - m * m1) ≠ m ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 19 from proof odd-mult (previous step) (back to tree)

2 * m0 = m * 2 * m1 + m, 2 * (m0 - m * m1) ≠ m ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: nat, name:elim-sub, seq:n ≤ m ⊦ n0 = m - n ↔ m = n0 + n, subst:[m, n, n0] → [m0, m * m1, n0]

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss) 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 ⊦ 2 * m = 2 * n + m0 → ¬ m < n; (s) n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0, m * m1, n0

Proof step 20 from proof odd-mult (previous step) (back to tree)

2 * m0 = m * 2 * m1 + m, m0 = n0 + m * m1, 2 * n0 ≠ m, m * m1 ≤ m0 ⊦

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 ⊦ m ≤ m + n; (s)

Proof step 21 from proof odd-mult (previous step) (back to tree)

2 * (n0 + m * m1) = m * 2 * m1 + m, 2 * n0 ≠ m ⊦

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] → [m * m1, n0, 2], <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 ⊦ m + n = m + n0 ↔ n = n0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: m * m1, n0, 2

Proof step 22 from proof odd-mult (previous step) (back to tree)

⊦ m0 * (m + n) = m0 * m + m0 * n

A spec-lemma

Proof Tree for odd-mult (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