Proof step 1 from proof gcd-a (conclusion) (back to tree)

⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0)

System applied rule normalize

And obtained the following new goals

Proof step 2:
gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0) ⊦

Rule argument

Additional information

Proof step 2 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0) ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 3:
m | n, n | m ⊦ m = n
Proof step 4:
gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0) ⊦ gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0) ∧ gcd(gcd(m, n), n0) | gcd(m, gcd(n, n0))

Rule argument

spec: <toplevel>, name:divi-eq, seq:m | n, n | m ⊦ m = n, subst:[m, n] → [gcd(m, gcd(n, n0)), gcd(gcd(m, n), n0)]

Additional information

Used terms: gcd(m, gcd(n, n0)), gcd(gcd(m, n), n0)

Proof step 3 from proof gcd-a (previous step) (back to tree)

m | n, n | m ⊦ m = n

Lemma divi-eq

Proof step 4 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0) ⊦ gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0) ∧ gcd(gcd(m, n), n0) | gcd(m, gcd(n, n0))

User applied rule case distinction

And obtained the following new goals

Proof step 5:
gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0), ¬ gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0) ⊦
Proof step 10:
gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0), ¬ gcd(gcd(m, n), n0) | gcd(m, gcd(n, n0)) ⊦

Rule argument

pos: right 1

Additional information

Proof step 5 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0), ¬ gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0) ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 6:
⊦ n0 | m ∧ n0 | n → n0 | gcd(m, n)
Proof step 7:
gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0), ¬ gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0), gcd(m, gcd(n, n0)) | gcd(m, n) ∧ gcd(m, gcd(n, n0)) | n0 → gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0) ⊦

Rule argument

spec: <toplevel>, name:gcd-alldiv, seq: ⊦ n0 | m ∧ n0 | n → n0 | gcd(m, n), subst:[n0, m, n] → [gcd(m, gcd(n, n0)), gcd(m, n), n0]

Additional information

Used terms: gcd(m, gcd(n, n0)), gcd(m, n), n0

Proof step 6 from proof gcd-a (previous step) (back to tree)

⊦ n0 | m ∧ n0 | n → n0 | gcd(m, n)

Lemma gcd-alldiv

Proof step 7 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0), ¬ gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0), gcd(m, gcd(n, n0)) | gcd(m, n) ∧ gcd(m, gcd(n, n0)) | n0 → gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0), ¬ gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0), ¬ gcd(m, gcd(n, n0)) | n0 ⊦

Rule argument

Additional information

Used simplifier rules: gcd-div-02: ⊦ gcd(m, gcd(n, n0)) | gcd(m, n);

Proof step 8 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0), ¬ gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0), ¬ gcd(m, gcd(n, n0)) | n0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ gcd(m, n) = gcd(n, m)

Rule argument

spec: <toplevel>, name:gcd-c, seq: ⊦ gcd(m, n) = gcd(n, m), subst:[n, m] → [n0, n], <all paths>

Additional information

Used simplifier rules: gcd-div-03: ⊦ gcd(m, gcd(n, n0)) | n; Used terms: n0, n

Proof step 9 from proof gcd-a (previous step) (back to tree)

⊦ gcd(m, n) = gcd(n, m)

Lemma gcd-c

Proof step 10 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0), ¬ gcd(gcd(m, n), n0) | gcd(m, gcd(n, n0)) ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 11:
⊦ n0 | m ∧ n0 | n → n0 | gcd(m, n)
Proof step 12:
gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0), ¬ gcd(gcd(m, n), n0) | gcd(m, gcd(n, n0)), gcd(gcd(m, n), n0) | m ∧ gcd(gcd(m, n), n0) | gcd(n, n0) → gcd(gcd(m, n), n0) | gcd(m, gcd(n, n0)) ⊦

Rule argument

spec: <toplevel>, name:gcd-alldiv, seq: ⊦ n0 | m ∧ n0 | n → n0 | gcd(m, n), subst:[n0, m, n] → [gcd(gcd(m, n), n0), m, gcd(n, n0)]

Additional information

Used terms: gcd(gcd(m, n), n0), m, gcd(n, n0)

Proof step 11 from proof gcd-a (previous step) (back to tree)

⊦ n0 | m ∧ n0 | n → n0 | gcd(m, n)

Lemma gcd-alldiv

Proof step 12 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(gcd(m, n), n0), ¬ gcd(gcd(m, n), n0) | gcd(m, gcd(n, n0)), gcd(gcd(m, n), n0) | m ∧ gcd(gcd(m, n), n0) | gcd(n, n0) → gcd(gcd(m, n), n0) | gcd(m, gcd(n, n0)) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ gcd(m, n) = gcd(n, m)
Proof step 14:
gcd(m, gcd(n, n0)) ≠ gcd(gcd(n, m), n0), ¬ gcd(gcd(n, m), n0) | gcd(m, gcd(n, n0)), gcd(gcd(n, m), n0) | m ∧ gcd(gcd(n, m), n0) | gcd(n, n0) → gcd(gcd(n, m), n0) | gcd(m, gcd(n, n0)) ⊦

Rule argument

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

Additional information

Used terms: n, m

Proof step 13 from proof gcd-a (previous step) (back to tree)

⊦ gcd(m, n) = gcd(n, m)

Lemma gcd-c

Proof step 14 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(gcd(n, m), n0), ¬ gcd(gcd(n, m), n0) | gcd(m, gcd(n, n0)), gcd(gcd(n, m), n0) | m ∧ gcd(gcd(n, m), n0) | gcd(n, n0) → gcd(gcd(n, m), n0) | gcd(m, gcd(n, n0)) ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 15:
gcd(m, gcd(n, n0)) ≠ gcd(gcd(n, m), n0), ¬ gcd(gcd(n, m), n0) | gcd(m, gcd(n, n0)) ⊦ gcd(gcd(n, m), n0) | m ∧ gcd(gcd(n, m), n0) | gcd(n, n0)

Rule argument

pos: left 3

Additional information

Proof step 15 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(gcd(n, m), n0), ¬ gcd(gcd(n, m), n0) | gcd(m, gcd(n, n0)) ⊦ gcd(gcd(n, m), n0) | m ∧ gcd(gcd(n, m), n0) | gcd(n, n0)

User applied rule apply lemma

And obtained the following new goals

Proof step 16:
⊦ gcd(m, gcd(n, n0)) | n
Proof step 17:
gcd(m, gcd(n, n0)) ≠ gcd(gcd(n, m), n0), gcd(n0, gcd(m, n)) | m, ¬ gcd(gcd(n, m), n0) | gcd(m, gcd(n, n0)) ⊦ gcd(gcd(n, m), n0) | m ∧ gcd(gcd(n, m), n0) | gcd(n, n0)

Rule argument

spec: <toplevel>, name:gcd-div-03, seq: ⊦ gcd(m, gcd(n, n0)) | n, subst:[m, n, n0] → [n0, m, n]

Additional information

Used terms: n0, m, n

Proof step 16 from proof gcd-a (previous step) (back to tree)

⊦ gcd(m, gcd(n, n0)) | n

Lemma gcd-div-03

Proof step 17 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(gcd(n, m), n0), gcd(n0, gcd(m, n)) | m, ¬ gcd(gcd(n, m), n0) | gcd(m, gcd(n, n0)) ⊦ gcd(gcd(n, m), n0) | m ∧ gcd(gcd(n, m), n0) | gcd(n, n0)

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 18:
⊦ gcd(m, n) = gcd(n, m)
Proof step 19:
gcd(m, gcd(n, n0)) ≠ gcd(n0, gcd(n, m)), gcd(n0, gcd(m, n)) | m, ¬ gcd(n0, gcd(n, m)) | gcd(m, gcd(n, n0)) ⊦ gcd(n0, gcd(n, m)) | m ∧ gcd(n0, gcd(n, m)) | gcd(n, n0)

Rule argument

spec: <toplevel>, name:gcd-c, seq: ⊦ gcd(m, n) = gcd(n, m), subst:[n, m] → [n0, gcd(n, m)], <all paths>

Additional information

Used terms: n0, gcd(n, m)

Proof step 18 from proof gcd-a (previous step) (back to tree)

⊦ gcd(m, n) = gcd(n, m)

Lemma gcd-c

Proof step 19 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(n0, gcd(n, m)), gcd(n0, gcd(m, n)) | m, ¬ gcd(n0, gcd(n, m)) | gcd(m, gcd(n, n0)) ⊦ gcd(n0, gcd(n, m)) | m ∧ gcd(n0, gcd(n, m)) | gcd(n, n0)

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ gcd(m, n) = gcd(n, m)
Proof step 21:
gcd(m, gcd(n, n0)) ≠ gcd(n0, gcd(m, n)), gcd(n0, gcd(m, n)) | m, ¬ gcd(n0, gcd(m, n)) | gcd(m, gcd(n, n0)) ⊦ gcd(n0, gcd(m, n)) | m ∧ gcd(n0, gcd(m, n)) | gcd(n, n0)

Rule argument

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

Additional information

Used terms: m, n

Proof step 20 from proof gcd-a (previous step) (back to tree)

⊦ gcd(m, n) = gcd(n, m)

Lemma gcd-c

Proof step 21 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(n0, gcd(m, n)), gcd(n0, gcd(m, n)) | m, ¬ gcd(n0, gcd(m, n)) | gcd(m, gcd(n, n0)) ⊦ gcd(n0, gcd(m, n)) | m ∧ gcd(n0, gcd(m, n)) | gcd(n, n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
gcd(m, gcd(n, n0)) ≠ gcd(n0, gcd(m, n)), ¬ gcd(n0, gcd(m, n)) | gcd(m, gcd(n, n0)), ¬ gcd(n0, gcd(m, n)) | gcd(n, n0) ⊦

Rule argument

Additional information

Used simplifier rules: gcd-div-03: ⊦ gcd(m, gcd(n, n0)) | n;

Proof step 22 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n, n0)) ≠ gcd(n0, gcd(m, n)), ¬ gcd(n0, gcd(m, n)) | gcd(m, gcd(n, n0)), ¬ gcd(n0, gcd(m, n)) | gcd(n, n0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ gcd(m, n) = gcd(n, m)
Proof step 24:
gcd(m, gcd(n0, n)) ≠ gcd(n0, gcd(m, n)), ¬ gcd(n0, gcd(m, n)) | gcd(m, gcd(n0, n)), ¬ gcd(n0, gcd(m, n)) | gcd(n0, n) ⊦

Rule argument

spec: <toplevel>, name:gcd-c, seq: ⊦ gcd(m, n) = gcd(n, m), subst:[n, m] → [n0, n], <all paths>

Additional information

Used terms: n0, n

Proof step 23 from proof gcd-a (previous step) (back to tree)

⊦ gcd(m, n) = gcd(n, m)

Lemma gcd-c

Proof step 24 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n0, n)) ≠ gcd(n0, gcd(m, n)), ¬ gcd(n0, gcd(m, n)) | gcd(m, gcd(n0, n)), ¬ gcd(n0, gcd(m, n)) | gcd(n0, n) ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 25:
⊦ gcd(m, gcd(n, n0)) | gcd(m, n)
Proof step 26:
gcd(m, gcd(n0, n)) ≠ gcd(n0, gcd(m, n)), gcd(n0, gcd(n, m)) | gcd(n0, n), ¬ gcd(n0, gcd(m, n)) | gcd(m, gcd(n0, n)), ¬ gcd(n0, gcd(m, n)) | gcd(n0, n) ⊦

Rule argument

spec: <toplevel>, name:gcd-div-02, seq: ⊦ gcd(m, gcd(n, n0)) | gcd(m, n), subst:[m, n, n0] → [n0, n, m]

Additional information

Used terms: n0, n, m

Proof step 25 from proof gcd-a (previous step) (back to tree)

⊦ gcd(m, gcd(n, n0)) | gcd(m, n)

Lemma gcd-div-02

Proof step 26 from proof gcd-a (previous step) (back to tree)

gcd(m, gcd(n0, n)) ≠ gcd(n0, gcd(m, n)), gcd(n0, gcd(n, m)) | gcd(n0, n), ¬ gcd(n0, gcd(m, n)) | gcd(m, gcd(n0, n)), ¬ gcd(n0, gcd(m, n)) | gcd(n0, n) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ gcd(m, n) = gcd(n, m)

Rule argument

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

Additional information

Used terms: m, n

Proof step 27 from proof gcd-a (previous step) (back to tree)

⊦ gcd(m, n) = gcd(n, m)

Lemma gcd-c

Proof Tree for gcd-a (View conclusion)

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