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Proof step 1 from proof abs-eq (conclusion) (back to tree)

⊦ i→n(i) = i→n(j) ↔ abs(i) = abs(j)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof abs-eq (previous step) (back to tree)

¬(i→n(i) = i→n(j) ↔ abs(i) = abs(j)) ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Proof step 3 from proof abs-eq (previous step) (back to tree)

i→n(i) = i→n(j), abs(i) ≠ abs(j) ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: int-basic1, name:tot, seq: ⊦ i < j ∨ i = j ∨ j < i, subst:[i, j] → [abs(i), abs(j)]

Additional information

Used terms: abs(i), abs(j)

Proof step 4 from proof abs-eq (previous step) (back to tree)

⊦ i < j ∨ i = j ∨ j < i

A spec-lemma

Proof step 5 from proof abs-eq (previous step) (back to tree)

i→n(i) = i→n(j), abs(i) ≠ abs(j), abs(i) < abs(j) ∨ abs(i) = abs(j) ∨ abs(j) < abs(i) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:rlabs, seq: ⊦ abs(i) < abs(j) ↔ i→n(i) < i→n(j), subst:[i, j] → [i, j], <all paths>

Additional information

Used terms: i, j

Proof step 6 from proof abs-eq (previous step) (back to tree)

⊦ abs(i) < abs(j) ↔ i→n(i) < i→n(j)

Lemma rlabs

Proof step 7 from proof abs-eq (previous step) (back to tree)

i→n(i) = i→n(j), abs(i) ≠ abs(j), i→n(i) < i→n(j) ∨ abs(i) = abs(j) ∨ abs(j) < abs(i) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:rlabs, seq: ⊦ abs(i) < abs(j) ↔ i→n(i) < i→n(j), subst:[j, i] → [i, j], <all paths>

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s) Used terms: i, j

Proof step 8 from proof abs-eq (previous step) (back to tree)

⊦ abs(i) < abs(j) ↔ i→n(i) < i→n(j)

Lemma rlabs

Proof step 9 from proof abs-eq (previous step) (back to tree)

abs(i) = abs(j), i→n(i) ≠ i→n(j) ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: nat-basic, name:tot, seq: ⊦ n < m ∨ n = m ∨ m < n, subst:[n, m] → [i→n(i), i→n(j)]

Additional information

Used terms: i→n(i), i→n(j)

Proof step 10 from proof abs-eq (previous step) (back to tree)

⊦ n < m ∨ n = m ∨ m < n

A spec-lemma

Proof step 11 from proof abs-eq (previous step) (back to tree)

abs(i) = abs(j), i→n(i) ≠ i→n(j), i→n(i) < i→n(j) ∨ i→n(i) = i→n(j) ∨ i→n(j) < i→n(i) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:rlabs, seq: ⊦ abs(i) < abs(j) ↔ i→n(i) < i→n(j), subst:[j, i] → [j, i], <rotated> <all paths>

Additional information

Used terms: j, i

Proof step 12 from proof abs-eq (previous step) (back to tree)

⊦ abs(i) < abs(j) ↔ i→n(i) < i→n(j)

Lemma rlabs

Proof step 13 from proof abs-eq (previous step) (back to tree)

abs(i) = abs(j), i→n(i) ≠ i→n(j), abs(i) < abs(j) ∨ i→n(i) = i→n(j) ∨ i→n(j) < i→n(i) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:rlabs, seq: ⊦ abs(i) < abs(j) ↔ i→n(i) < i→n(j), subst:[j, i] → [i, j], <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ ¬ i < i; (s) Used terms: i, j

Proof step 14 from proof abs-eq (previous step) (back to tree)

⊦ abs(i) < abs(j) ↔ i→n(i) < i→n(j)

Lemma rlabs

Proof Tree for abs-eq (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