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

n ≤ m ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0

User applied rule induction

And obtained the following new goals

Proof step 2:
n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0

Rule argument

indvar: m wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: m

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

n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0

User applied rule cut formula

And obtained the following new goals

Proof step 3:
n = 0, n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0
Proof step 5:
n ≠ 0, n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0

Rule argument

fma: n = 0

Additional information

Used formulas: n = 0

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

n = 0, n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0. m * m0 + m = 0 ∨ m = 0 ∨ m0 = 1

Rule argument

Additional information

Used simplifier rules: gcd-zero-02: ⊦ gcd(n, 0) = n; gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0); mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult ⊦ m * n = m ↔ m = 0 ∨ n = 1; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ 0 + m = m; (s) ⊦ m * 0 = 0; (s) ⊦ 0 ≤ m; (s)

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

∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0. m * m0 + m = 0 ∨ m = 0 ∨ m0 = 1

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: 1 abort: F discard: T computed: F

Additional information

Used terms: 1

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

n ≠ 0, n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 6:
n ≠ 0 ⊦ gcd(m, n) = gcd(n, m % n)
Proof step 7:
n ≠ 0, n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

Rule argument

spec: <toplevel>, name:gcd-mod, seq:n ≠ 0 ⊦ gcd(m, n) = gcd(n, m % n), subst:[n, m] → [n, m], <all paths>

Additional information

Used terms: n, m

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

n ≠ 0 ⊦ gcd(m, n) = gcd(n, m % n)

Lemma gcd-mod

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

n ≠ 0, n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n ≠ 0, n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

Rule argument

Additional information

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

User applied rule cut formula

And obtained the following new goals

Proof step 9:
m = n, n ≠ 0, n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0
Proof step 11:
m ≠ n, n ≠ 0, n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

Rule argument

fma: m = n

Additional information

Used formulas: m = n

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

m = n, n ≠ 0, n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
n ≠ 0, ∀ m0, n0. m0 < n → n0 ≤ m0 → (∃ m1, n1. m0 * m1 + gcd(m0, n0) = n0 * n1 ∨ n0 * n1 + gcd(m0, n0) = m0 * m1) ⊦ ∃ m0, n0. m0 + 1 = n0 ∨ n0 + 1 = m0

Rule argument

Additional information

Used simplifier rules: gcd-zero-02: ⊦ gcd(n, 0) = n; gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0); 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 * n + n = m1 * n ↔ m + 1 = m1 ∨ n = 0; (s) n ≠ 0 ⊦ n % n = 0; (s) ⊦ m ≤ m; (s)

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

n ≠ 0, ∀ m0, n0. m0 < n → n0 ≤ m0 → (∃ m1, n1. m0 * m1 + gcd(m0, n0) = n0 * n1 ∨ n0 * n1 + gcd(m0, n0) = m0 * m1) ⊦ ∃ m0, n0. m0 + 1 = n0 ∨ n0 + 1 = m0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: 0, 1 abort: F discard: T computed: F

Additional information

Used terms: 0, 1

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

m ≠ n, n ≠ 0, n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m ≠ n, n ≠ 0, n < m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)

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

m ≠ n, n ≠ 0, n < m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

User applied rule apply induction

And obtained the following new goals

Proof step 13:
m ≠ n, n ≠ 0, n < m, ∃ m1, n0. n * m1 + gcd(n, m % n) = m % n * n0 ∨ m % n * n0 + gcd(n, m % n) = n * m1, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

Rule argument

subst: [m0, n] → [n, m % n]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m % n ≤ n; (s) Used terms: n, m % n

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

m ≠ n, n ≠ 0, n < m, ∃ m1, n0. n * m1 + gcd(n, m % n) = m % n * n0 ∨ m % n * n0 + gcd(n, m % n) = n * m1, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
m ≠ n, n ≠ 0, n < m, n * m1 + gcd(n, m % n) = m % n * n0 ∨ m % n * n0 + gcd(n, m % n) = n * m1, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

Rule argument

Additional information

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

m ≠ n, n ≠ 0, n < m, n * m1 + gcd(n, m % n) = m % n * n0 ∨ m % n * n0 + gcd(n, m % n) = n * m1, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ n ≠ 0 → m % n = m - m / n * n
Proof step 16:
m ≠ n, n ≠ 0, n < m, n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0 ∨ (m - m / n * n) * n0 + gcd(n, m - m / n * n) = n * m1, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m - m / n * n) = n * n0 ∨ n * n0 + gcd(n, m - m / n * n) = m * m0

Rule argument

spec: nat-div, name:mod-rew, seq: ⊦ n ≠ 0 → m % n = m - m / n * n, subst:[n, m] → [n, m], <all paths>

Additional information

Used terms: n, m

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

⊦ n ≠ 0 → m % n = m - m / n * n

A spec-lemma

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

m ≠ n, n ≠ 0, n < m, n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0 ∨ (m - m / n * n) * n0 + gcd(n, m - m / n * n) = n * m1, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m - m / n * n) = n * n0 ∨ n * n0 + gcd(n, m - m / n * n) = m * m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
m ≠ n, n ≠ 0, n < m, n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0 ∨ (m - m / n * n) * n0 + gcd(n, m - m / n * n) = n * m1, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(n, m - m / n * n) = n * n0 ∨ n * n0 + gcd(n, m - m / n * n) = m * m0

Rule argument

Additional information

Proof step 17 from proof gcdexh (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

Proof step 18:
m * n0 + gcd(n, m - m / n * n) ≠ n * (m1 + m / n * n0), n * (m1 + m / n * n0) + gcd(n, m - m / n * n) ≠ m * n0, m ≠ n, n ≠ 0, n < m, n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0 ∨ (m - m / n * n) * n0 + gcd(n, m - m / n * n) = n * m1, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦

Rule argument

pos: right 1 terms: n0, m1 + m / n * n0 abort: F discard: T computed: F

Additional information

Used terms: n0, m1 + m / n * n0

Proof step 18 from proof gcdexh (previous step) (back to tree)

m * n0 + gcd(n, m - m / n * n) ≠ n * (m1 + m / n * n0), n * (m1 + m / n * n0) + gcd(n, m - m / n * n) ≠ m * n0, m ≠ n, n ≠ 0, n < m, n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0 ∨ (m - m / n * n) * n0 + gcd(n, m - m / n * n) = n * m1, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
m * n0 + gcd(n, m - m / n * n) ≠ n * (m1 + m / n * n0), n * (m1 + m / n * n0) + gcd(n, m - m / n * n) ≠ m * n0, m ≠ n, n ≠ 0, n < m, n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0 ∨ (m - m / n * n) * n0 + gcd(n, m - m / n * n) = n * m1, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦

Rule argument

Additional information

Proof step 19 from proof gcdexh (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ m0 * (m + n) = m0 * m + m0 * n
Proof step 21:
m * n0 + gcd(n, m - m / n * n) ≠ n * m1 + n * m / n * n0, (n * m1 + n * m / n * n0) + gcd(n, m - m / n * n) ≠ m * n0, m ≠ n, n ≠ 0, n < m, n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0 ∨ (m - m / n * n) * n0 + gcd(n, m - m / n * n) = n * m1, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦

Rule argument

spec: nat-mult, name:dist, seq: ⊦ m0 * (m + n) = m0 * m + m0 * n, subst:[n, m, m0] → [m / n * n0, m1, n], <all paths>

Additional information

Used terms: m / n * n0, m1, n

Proof step 20 from proof gcdexh (previous step) (back to tree)

⊦ m0 * (m + n) = m0 * m + m0 * n

A spec-lemma

Proof step 21 from proof gcdexh (previous step) (back to tree)

m * n0 + gcd(n, m - m / n * n) ≠ n * m1 + n * m / n * n0, (n * m1 + n * m / n * n0) + gcd(n, m - m / n * n) ≠ m * n0, m ≠ n, n ≠ 0, n < m, n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0 ∨ (m - m / n * n) * n0 + gcd(n, m - m / n * n) = n * m1, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 22:
n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0, m * n0 + gcd(n, m - m / n * n) ≠ n * m1 + n * m / n * n0, (n * m1 + n * m / n * n0) + gcd(n, m - m / n * n) ≠ m * n0, m ≠ n, n ≠ 0, n < m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦
Proof step 28:
(m - m / n * n) * n0 + gcd(n, m - m / n * n) = n * m1, m * n0 + gcd(n, m - m / n * n) ≠ n * m1 + n * m / n * n0, (n * m1 + n * m / n * n0) + gcd(n, m - m / n * n) ≠ m * n0, m ≠ n, n ≠ 0, n < m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦

Rule argument

pos: left 6

Additional information

Proof step 22 from proof gcdexh (previous step) (back to tree)

n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0, m * n0 + gcd(n, m - m / n * n) ≠ n * m1 + n * m / n * n0, (n * m1 + n * m / n * n0) + gcd(n, m - m / n * n) ≠ m * n0, m ≠ n, n ≠ 0, n < m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ (m + n0) + n1 = m + n0 + n1
Proof step 24:
n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0, m * n0 + gcd(n, m - m / n * n) ≠ n * m1 + n * m / n * n0, n * m1 + n * m / n * n0 + gcd(n, m - m / n * n) ≠ m * n0, m ≠ n, n ≠ 0, n < m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦

Rule argument

spec: nat-basic, name:a, seq: ⊦ (m + n0) + n1 = m + n0 + n1, subst:[n1, n0, m] → [gcd(n, m - m / n * n), n * m / n * n0, n * m1], <all paths>

Additional information

Used terms: gcd(n, m - m / n * n), n * m / n * n0, n * m1

Proof step 23 from proof gcdexh (previous step) (back to tree)

⊦ (m + n0) + n1 = m + n0 + n1

A spec-lemma

Proof step 24 from proof gcdexh (previous step) (back to tree)

n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0, m * n0 + gcd(n, m - m / n * n) ≠ n * m1 + n * m / n * n0, n * m1 + n * m / n * n0 + gcd(n, m - m / n * n) ≠ m * n0, m ≠ n, n ≠ 0, n < m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ m + n = n + m
Proof step 26:
n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0, m * n0 + gcd(n, m - m / n * n) ≠ n * m1 + n * m / n * n0, n * m1 + gcd(n, m - m / n * n) + n * m / n * n0 ≠ m * n0, m ≠ n, n ≠ 0, n < m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦

Rule argument

spec: nat-basic, name:c, seq: ⊦ m + n = n + m, subst:[n, m] → [gcd(n, m - m / n * n), n * m / n * n0], <all paths>

Additional information

Used terms: gcd(n, m - m / n * n), n * m / n * n0

Proof step 25 from proof gcdexh (previous step) (back to tree)

⊦ m + n = n + m

A spec-lemma

Proof step 26 from proof gcdexh (previous step) (back to tree)

n * m1 + gcd(n, m - m / n * n) = (m - m / n * n) * n0, m * n0 + gcd(n, m - m / n * n) ≠ n * m1 + n * m / n * n0, n * m1 + gcd(n, m - m / n * n) + n * m / n * n0 ≠ m * n0, m ≠ n, n ≠ 0, n < m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ (m + n0) + n1 = m + n0 + n1

Rule argument

spec: nat-basic, name:a, seq: ⊦ (m + n0) + n1 = m + n0 + n1, subst:[n1, n0, m] → [n * m / n * n0, gcd(n, m - m / n * n), n * m1], <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ m * n + m0 * n = m1 * n ↔ m + m0 = m1 ∨ n = 0; (s) n ≠ 0 ⊦ m / n * n ≤ m; (s) n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (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: n * m / n * n0, gcd(n, m - m / n * n), n * m1

Proof step 27 from proof gcdexh (previous step) (back to tree)

⊦ (m + n0) + n1 = m + n0 + n1

A spec-lemma

Proof step 28 from proof gcdexh (previous step) (back to tree)

(m - m / n * n) * n0 + gcd(n, m - m / n * n) = n * m1, m * n0 + gcd(n, m - m / n * n) ≠ n * m1 + n * m / n * n0, (n * m1 + n * m / n * n0) + gcd(n, m - m / n * n) ≠ m * n0, m ≠ n, n ≠ 0, n < m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 29:
⊦ m0 ≤ m → (m - m0) * n = m * n - m0 * n

Rule argument

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

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n; (ws) n ≠ 0 ⊦ ¬ m < m / n * n; (s) mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic n ≠ 0 ⊦ m / n * n ≤ m; (s) Used terms: n0, m / n * n, m

Proof step 29 from proof gcdexh (previous step) (back to tree)

⊦ m0 ≤ m → (m - m0) * n = m * n - m0 * n

A spec-lemma

Proof Tree for gcdexh (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.