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Proof step 1 from proof exgcd-div (conclusion) (back to tree)

⊦ ∃ i0. i / gcd(i, j) = i0 ∧ gcd(i, j) * i0 = i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof exgcd-div (previous step) (back to tree)

gcd(i, j) * i / gcd(i, j) ≠ i ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:gcd-div1, seq: ⊦ gcd(i, j) | i, subst:[i, j] → [i, j]

Additional information

Used terms: i, j

Proof step 3 from proof exgcd-div (previous step) (back to tree)

⊦ gcd(i, j) | i

Lemma gcd-div1

Proof step 4 from proof exgcd-div (previous step) (back to tree)

gcd(i, j) * i / gcd(i, j) ≠ i, gcd(i, j) | i ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:divides, seq: ⊦ i | j ↔ (∃ i0. i * i0 = j), subst:[j, i] → [i, gcd(i, j)], <all paths>

Additional information

Used terms: i, gcd(i, j)

Proof step 5 from proof exgcd-div (previous step) (back to tree)

⊦ i | j ↔ (∃ i0. i * i0 = j)

Lemma divides

Proof step 6 from proof exgcd-div (previous step) (back to tree)

gcd(i, j) * i / gcd(i, j) ≠ i, ∃ i0. gcd(i, j) * i0 = i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 7 from proof exgcd-div (previous step) (back to tree)

gcd(i, j) * i0 = i, gcd(i, j) * i / gcd(i, j) ≠ i ⊦

User applied rule subst equation

And obtained the following new goals

Rule argument

pos: left 1, flags: rotate, [[1,2,1,1,2,1]]

Additional information

Proof step 8 from proof exgcd-div (previous step) (back to tree)

gcd(i, j) * i0 = i, gcd(i, j) * (gcd(i, j) * i0) / gcd(i, j) ≠ i ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div, name:div-mult, seq:j ≠ 0 ⊦ (j * i) / j = i, subst:[i, j] → [i0, gcd(i, j)], <all paths>

Additional information

Used simplifier rules: gcd-zero-zero: ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0; gcd-zero: ⊦ gcd(0, 0) = 0; ⊦ i * 0 = 0; (s) associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: i0, gcd(i, j)

Proof step 9 from proof exgcd-div (previous step) (back to tree)

j ≠ 0 ⊦ (j * i) / j = i

A spec-lemma

Proof Tree for exgcd-div (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