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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(ins≤(a, x) = ins≤(a, y) ↔ x = y) ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ins≤(a, x) = ins≤(a, y), x ≠ y ⊦

Rule argument

Additional information

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

User applied rule structural induction

And obtained the following new goals

• Proof step 4:

  ins≤(a, []) = ins≤(a, y), [] ≠ y ⊦

• Proof step 8:

  ins≤(a, a0 + x0) = ins≤(a, y), a0 + x0 ≠ y, ∀ a, y. ¬(ins≤(a, x0) = ins≤(a, y) ∧ x0 ≠ y) ⊦

Rule argument

Additional information

Used simplifier rules: list freely generated by [], +;;

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  a ' = ins≤(a, y), [] ≠ y ⊦

Rule argument

Additional information

Used simplifier rules: ins-e: ⊦ ins≤(a, []) = a ';

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

User applied rule structural induction

And obtained the following new goals

• Proof step 6:

  a ' = ins≤(a, a0 + y0), [] ≠ a0 + y0, ∀ a. ¬(a ' = ins≤(a, y0) ∧ [] ≠ y0) ⊦

Rule argument

Additional information

Used simplifier rules: list freely generated by [], +;;

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  a ' = ins≤(a, a0 ' + y0), (∀ a1. a1 ' ≠ ins≤(a1, y0)) ∨ [] = y0 ⊦

Rule argument

Additional information

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

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

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ins-n: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x); empty: ⊦ ins≤(a, x) ≠ []; ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; ⊦ a + x = a ' + x; (s) ⊦ a ' = b ' + x ↔ a = b ∧ x = []; (ws) ⊦ a ' + x ≠ []; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used formulas: a0 < a

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  ins≤(a, a0 ' + x0) = ins≤(a, y), a0 ' + x0 ≠ y, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

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

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

• Proof step 10:

  ins≤(a, a0 ' + x0) = ins≤(a, y), a0 ' + x0 ≠ y, a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

• Proof step 19:

  ins≤(a, a0 ' + x0) = ins≤(a, y), a0 ' + x0 ≠ y, ¬ a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used formulas: a0 < a

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  a0 ' + ins≤(a, x0) = ins≤(a, y), a0 ' + x0 ≠ y, a0 < a, a0 ∈ ins≤(a, y), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

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 in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list ⊦ a + x = a ' + x; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 12:

  ⊦ x ≠ [] ∧ x.head < a → ins≤(a, x) = x.head + ins≤(a, x.tail)

• Proof step 13:

  a0 ' + ins≤(a, x0) = ins≤(a, y), a0 ' + x0 ≠ y, a0 < a, a0 ∈ ins≤(a, y), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦ y ≠ [] ∧ y.head < a

• Proof step 15:

  a0 ' + ins≤(a, x0) = y.head + ins≤(a, y.tail), a0 ' + x0 ≠ y, y ≠ [], a0 < a, y.head < a, a0 ∈ y.head + ins≤(a, y.tail), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used terms: y, a

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

Lemma ins-rec

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  a0 ' + ins≤(a, x0) = ins≤(a, y), a0 ' + x0 ≠ y, y ≠ [], a0 < a, a0 ∈ ins≤(a, y), ¬ y.head < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used simplifier rules: ins-e: ⊦ ins≤(a, []) = a '; empty: ⊦ ins≤(a, x) ≠ []; a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ' = b ' + x ↔ a = b ∧ x = []; (ws)

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

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws) ⊦ a ' + x ≠ []; (s) ⊦ a + x = a ' + x; (s) ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s) ⊦ a ∈ b ' ↔ a = b; (s) in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list a: ⊦ (x + y) + z = x + y + z; from specification list lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem ⊦ ¬ a < a; (s) x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: y, a2, y1

Proof step 15 from proof eq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  ins≤(a, x0) = ins≤(a, y.tail), y.head ' + x0 ≠ y, y ≠ [], y.head < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used simplifier rules: lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ b ' ↔ a = b; (s) ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s) ⊦ a + x = a ' + x; (s) ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws)

Proof step 16 from proof eq (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 17:

  ins≤(a, x0) = ins≤(a, y1), y = a0 + y1, a0 ' + x0 ≠ y, y ≠ [], a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: y, a0, y1

Proof step 17 from proof eq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  ins≤(a, x0) = ins≤(a, y1), x0 ≠ y1, a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

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

Proof step 18 from proof eq (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: a, y1

Proof step 19 from proof eq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  a ' + a0 ' + x0 = ins≤(a, y), a0 ' + x0 ≠ y, ¬ a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; in-04: ⊦ a ∈ 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 20 from proof eq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 21:

  ⊦ x ≠ [] ∧ x.head < a → ins≤(a, x) = x.head + ins≤(a, x.tail)

• Proof step 22:

  a ' + a0 ' + x0 = ins≤(a, y), a0 ' + x0 ≠ y, ¬ a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦ y ≠ [] ∧ y.head < a

Rule argument

Additional information

Used simplifier rules: ⊦ 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 a: ⊦ (x + y) + z = x + y + z; from specification list nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem ⊦ ¬ a < a; (s) Used terms: y, a

Proof step 21 from proof eq (previous step) (back to tree)

Lemma ins-rec

Proof step 22 from proof eq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  a ' + a0 ' + x0 = ins≤(a, y), a0 ' + x0 ≠ y, y ≠ [], ¬ a0 < a, ¬ y.head < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used simplifier rules: ins-e: ⊦ ins≤(a, []) = a '; a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ' = b ' + x ↔ a = b ∧ x = []; (ws)

Proof step 23 from proof eq (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; ⊦ a + x = a ' + x; (s) ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws) a: ⊦ (x + y) + z = x + y + z; from specification list x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: y, a2, y1