Proof step 1 from proof ole-01 (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(ordered≤(ins≤(a, x)) ↔ ordered≤(x)) ⊦

Rule argument

Additional information

Proof step 2 from proof ole-01 (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 3:

  ¬(ordered≤(ins≤(a, a0 + x0)) ↔ ordered≤(a0 + x0)), ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: le-e: ⊦ ordered≤([]); ins-e: ⊦ ins≤(a, []) = a '; le-o: ⊦ ordered≤(a '); list freely generated by [], +;;

Proof step 3 from proof ole-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 4:

  ¬(ordered≤(ins≤(a, a0 ' + x0)) ↔ ordered≤(a0 ' + x0)), ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)

Proof step 4 from proof ole-01 (previous step) (back to tree)

Heuristic weak cut applied rule cut formula

And obtained the following new goals

• Proof step 5:

  ¬(ordered≤(ins≤(a, a0 ' + x0)) ↔ ordered≤(a0 ' + x0)), a0 < a, ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: app-01: ⊦ ordered≤(x + y) → ordered≤(y); ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; le-r: ⊦ ordered≤(a ' + b ' + x) ↔ ¬ b < a ∧ ordered≤(b ' + x); ⊦ a + x = a ' + x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used formulas: a0 < a

Proof step 5 from proof ole-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  ¬(ordered≤(a0 ' + ins≤(a, x0)) ↔ ordered≤(a0 ' + x0)), a0 < a, ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: ins-n: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x); a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a + x = a ' + x; (s)

Proof step 6 from proof ole-01 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ x ≠ [] ∧ x.tail ≠ [] → (ordered≤(x) ↔ ¬ x.tail.head < x.head ∧ ordered≤(x.tail))

• Proof step 8:

  a0 ' + ins≤(a, x0) ≠ [], (a0 ' + ins≤(a, x0)).tail ≠ [], ¬(¬ (a0 ' + ins≤(a, x0)).tail.head < (a0 ' + ins≤(a, x0)).head ∧ ordered≤((a0 ' + ins≤(a, x0)).tail) ↔ ordered≤(a0 ' + x0)), a0 < a, ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: empty: ⊦ ins≤(a, x) ≠ []; ⊦ a ' + x ≠ []; (s) ⊦ (a ' + x).tail = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a0 ' + ins≤(a, x0)

Proof step 7 from proof ole-01 (previous step) (back to tree)

Lemma ole-rec

Proof step 8 from proof ole-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  ¬(¬ ins≤(a, x0).head < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ordered≤(a0 ' + x0)), a0 < a, ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: empty: ⊦ ins≤(a, x) ≠ []; a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ (a ' + x).tail = x; (s) ⊦ (a ' + x).head = a; (s) ⊦ a ' + x ≠ []; (s)

Proof step 9 from proof ole-01 (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 10:

  ins≤(a, x0) = a1 + y, ¬(¬ a1 < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ordered≤(a0 ' + x0)), ins≤(a, x0) ≠ [], a0 < a, ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: empty: ⊦ ins≤(a, x) ≠ []; x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: ins≤(a, x0), a1, y

Proof step 10 from proof ole-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  ins≤(a, x0) = a1 ' + y, ¬(¬ a1 < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ordered≤(a0 ' + x0)), a0 < a, a1 ∈ ins≤(a, x0), ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: empty: ⊦ ins≤(a, x) ≠ []; a: ⊦ (x + y) + z = x + y + z; from specification list in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list ⊦ a + x = a ' + x; (s)

Proof step 11 from proof ole-01 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 12:

  ⊦ x ≠ [] ∧ x.tail ≠ [] → (ordered≤(x) ↔ ¬ x.tail.head < x.head ∧ ordered≤(x.tail))

• Proof step 13:

  ins≤(a, x0) = a1 ' + y, a0 ' + x0 ≠ [], (a0 ' + x0).tail ≠ [], ¬(¬ a1 < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ¬ (a0 ' + x0).tail.head < (a0 ' + x0).head ∧ ordered≤((a0 ' + x0).tail)), a0 < a, a1 ∈ ins≤(a, x0), ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: le-e: ⊦ ordered≤([]); ins-e: ⊦ ins≤(a, []) = a '; le-o: ⊦ ordered≤(a '); ⊦ a ' + x ≠ []; (s) ⊦ (a ' + x).tail = x; (s) ⊦ a ' = b ' + x ↔ a = b ∧ x = []; (ws) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ b ' ↔ a = b; (s) transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem ⊦ ¬ a < a; (s) Used terms: a0 ' + x0

Proof step 12 from proof ole-01 (previous step) (back to tree)

Lemma ole-rec

Proof step 13 from proof ole-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  ins≤(a, x0) = a1 ' + y, ¬(¬ a1 < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ¬ x0.head < a0 ∧ ordered≤(x0)), x0 ≠ [], a0 < a, a1 ∈ ins≤(a, x0), ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ (a ' + x).tail = x; (s) ⊦ (a ' + x).head = a; (s) ⊦ a ' + x ≠ []; (s)

Proof step 14 from proof ole-01 (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 15:

  ins≤(a, x0) = a1 ' + y, x0 = a2 + y0, ¬(¬ a1 < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ¬ a2 < a0 ∧ ordered≤(x0)), x0 ≠ [], a0 < a, a1 ∈ ins≤(a, x0), ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x0, a2, y0

Proof step 15 from proof ole-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  ins≤(a, a2 ' + y0) = a1 ' + y, ¬(¬ a1 < a0 ∧ ordered≤(a1 ' + y) ↔ ¬ a2 < a0 ∧ ordered≤(a2 ' + y0)), a0 < a, ∀ a3. ordered≤(ins≤(a3, a2 ' + y0)) ↔ ordered≤(a2 ' + y0) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ b ' + x ↔ a = b ∨ a ∈ x; (s) in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list ⊦ a + x = a ' + x; (s) ⊦ a ' + x ≠ []; (s)

Proof step 16 from proof ole-01 (previous step) (back to tree)

Heuristic weak cut applied rule cut formula

And obtained the following new goals

• Proof step 17:

  ins≤(a, a2 ' + y0) = a1 ' + y, ¬(¬ a1 < a0 ∧ ordered≤(a1 ' + y) ↔ ¬ a2 < a0 ∧ ordered≤(a2 ' + y0)), a0 < a, a2 < a, ∀ a3. ordered≤(ins≤(a3, a2 ' + y0)) ↔ ordered≤(a2 ' + y0) ⊦

Rule argument

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; le-r: ⊦ ordered≤(a ' + b ' + x) ↔ ¬ b < a ∧ ordered≤(b ' + x); cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x); app-01: ⊦ ordered≤(x + y) → ordered≤(y); ⊦ a + x = a ' + x; (s) ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws) in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem ⊦ ¬ a < a; (s) ⊦ a ∈ b ' + x ↔ a = b ∨ a ∈ x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem Used formulas: a2 < a

Proof step 17 from proof ole-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  ¬(ordered≤(a1 ' + ins≤(a, y0)) ↔ ordered≤(a1 ' + y0)), a1 < a, a0 < a, ¬ a1 < a0, ∀ a3. ordered≤(ins≤(a3, a1 ' + y0)) ↔ ordered≤(a1 ' + y0) ⊦

Rule argument

Additional information

Used simplifier rules: ins-n: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x); nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws) ⊦ a + x = a ' + x; (s)

Proof step 18 from proof ole-01 (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ins-n: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x); ⊦ a + x = a ' + x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a