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Proof step 1 from proof ordered-arraylist-append-last (conclusion) (back to tree)

⊦ ordered≤(arx + ar0 ') → (ordered≤(arx + ar0 ' + ar ') ↔ ar0 ≤ ar)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof ordered-arraylist-append-last (previous step) (back to tree)

¬(ordered≤(arx + ar0 ' + ar ') ↔ ar0 ≤ ar), ordered≤(arx + ar0 ') ⊦

User applied rule structural induction

And obtained the following new goals

Rule argument

term: arx

Additional information

Used simplifier rules: ordered-arraylist-two: ⊦ ordered≤(ar0 ' + ar1 ') ↔ ar0 ≤ ar1; list-of-natarray freely generated by [], +;; ⊦ [] + x = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[list-of-natarray]

Proof step 3 from proof ordered-arraylist-append-last (previous step) (back to tree)

¬(ordered≤((ar1 + arx0) + ar0 ' + ar ') ↔ ar0 ≤ ar), ordered≤((ar1 + arx0) + ar0 '), ∀ ar0, ar. ¬(¬(ordered≤(arx0 + ar0 ' + ar ') ↔ ar0 ≤ ar) ∧ ordered≤(arx0 + ar0 ')) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list[list-of-natarray] ⊦ a + x = a ' + x; (s)

Proof step 4 from proof ordered-arraylist-append-last (previous step) (back to tree)

¬(ordered≤((ar1 ' + arx0) + ar0 ' + ar ') ↔ ar0 ≤ ar), ordered≤((ar1 ' + arx0) + ar0 '), ∀ ar2, ar3. (ordered≤(arx0 + ar2 ' + ar3 ') ↔ ar2 ≤ ar3) ∨ ¬ ordered≤(arx0 + ar2 ') ⊦

User applied rule constructor cut

And obtained the following new goals

Rule argument

var: arx0

Additional information

Used simplifier rules: ordered-arraylist-rec: ⊦ ordered≤(ar0 ' + ar1 ' + arx) ↔ ar0 ≤ ar1 ∧ ordered≤(ar1 ' + arx); ordered-arraylist-two: ⊦ ordered≤(ar0 ' + ar1 ') ↔ ar0 ≤ ar1; ordered-arraylist-base1: ⊦ ordered≤(ar '); list-of-natarray freely generated by [], +;; ⊦ [] + x = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[list-of-natarray]

Proof step 5 from proof ordered-arraylist-append-last (previous step) (back to tree)

arx0 = ar4 + arx, ¬(ordered≤((ar1 ' + arx0) + ar0 ' + ar ') ↔ ar0 ≤ ar), ordered≤((ar1 ' + arx0) + ar0 '), ∀ ar2, ar3. (ordered≤(arx0 + ar2 ' + ar3 ') ↔ ar2 ≤ ar3) ∨ ¬ ordered≤(arx0 + ar2 ') ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ordered-arraylist-rec: ⊦ ordered≤(ar0 ' + ar1 ' + arx) ↔ ar0 ≤ ar1 ∧ ordered≤(ar1 ' + arx); a: ⊦ (x + y) + z = x + y + z; from specification list[list-of-natarray] ⊦ a + x = a ' + x; (s)

Proof step 6 from proof ordered-arraylist-append-last (previous step) (back to tree)

¬(ordered≤(ar4 ' + arx + ar0 ' + ar ') ↔ ar0 ≤ ar), ordered≤(ar4 ' + arx + ar0 '), ar1 ≤ ar4, ∀ ar2, ar3. (ordered≤((ar4 ' + arx) + ar2 ' + ar3 ') ↔ ar2 ≤ ar3) ∨ ¬ ordered≤((ar4 ' + arx) + ar2 ') ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: ar0, ar abort: F discard: T computed: T

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list[list-of-natarray] Used terms: ar0, ar

Proof Tree for ordered-arraylist-append-last (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