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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  # ax ≠ # ay, unique_al(ay), unique_al(ax), ∀ a. a ∈ ax ↔ a ∈ ay ⊦

Rule argument

Additional information

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

User applied rule structural induction

And obtained the following new goals

• Proof step 3:

  # [] ≠ # ay, unique_al([]), unique_al(ay), ∀ a. a ∈ [] ↔ a ∈ ay ⊦

• Proof step 8:

  #(pair + ax0) ≠ # ay, unique_al(pair + ax0), unique_al(ay), ∀ a. a ∈ pair + ax0 ↔ a ∈ ay, ∀ ay. ¬(# ax0 ≠ # ay ∧ unique_al(ay) ∧ unique_al(ax0) ∧ (∀ a. a ∈ ax0 ↔ a ∈ ay)) ⊦

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 4:

  ay ≠ [], unique_al(ay), ∀ a. ¬ a ∈ ay ⊦

Rule argument

Additional information

Used simplifier rules: unique-e: ⊦ unique_al([]); empty: ⊦ ¬ a ∈ []; ⊦ # x = 0 ↔ x = []; (s) ⊦ # [] = 0; (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 5:

  ay ≠ [], unique_al(ay), ¬ ay.head.1 ∈ ay, ∀ a. ¬ a ∈ ay ⊦

Rule argument

Additional information

Used terms: ay.head.1

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 6:

  ay = head + ay0, ay ≠ [], unique_al(ay), ¬ head.1 ∈ ay, ∀ a. ¬ a ∈ ay ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ x ≠ [] → (head = x.head ∧ tail = x.tail ↔ x = head + tail); (elim) Used terms: ay, head, tail

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  unique_al(head ' + ay0), ¬ head.1 ∈ head ', ¬ head.1 ∈ ay0, ∀ a. ¬ a ∈ head ' ∧ ¬ a ∈ ay0 ⊦

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ a ∈ ax ∨ a ∈ ay; a: ⊦ (x + y) + z = x + y + z; from specification list[assoclist] ⊦ a + x = a ' + x; (s) ⊦ a ' + x ≠ []; (s)

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

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

Additional information

Used simplifier rules: cons: ⊦ unique_al((a × d)' + ax) ↔ ¬ a ∈ ax ∧ unique_al(ax); inone: ⊦ a ∈ (b × d0)' ↔ a = b; a: ⊦ (x + y) + z = x + y + z; from specification list[assoclist] ⊦ a = p.1 ∧ d = p.2 ↔ p = a × d; (elim) Used terms: head, a0, d

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  # ax0 + 1 ≠ # ay, unique_al(ay), unique_al(pair ' + ax0), ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬ unique_al(ax0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0), ∀ a. a ∈ pair ' ∨ a ∈ ax0 ↔ a ∈ ay ⊦

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ a ∈ ax ∨ a ∈ ay; a: ⊦ (x + y) + z = x + y + z; from specification list[assoclist] ⊦ a + x = a ' + x; (s) ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s)

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

User applied rule structural induction

And obtained the following new goals

• Proof step 10:

  # ax0 + 1 ≠ # ay, unique_al((a × d)' + ax0), unique_al(ay), ∀ a0. a0 ∈ (a × d)' ∨ a0 ∈ ax0 ↔ a0 ∈ ay, ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬ unique_al(ax0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0) ⊦

Rule argument

Additional information

Used simplifier rules: pair freely generated by ×;;

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  # ax0 + 1 ≠ # ay, unique_al(ay), unique_al(ax0), ¬ a ∈ ax0, ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0), ∀ a0. a0 = a ∨ a0 ∈ ax0 ↔ a0 ∈ ay ⊦

Rule argument

Additional information

Used simplifier rules: cons: ⊦ unique_al((a × d)' + ax) ↔ ¬ a ∈ ax ∧ unique_al(ax); inone: ⊦ a ∈ (b × d0)' ↔ a = b; a: ⊦ (x + y) + z = x + y + z; from specification list[assoclist]

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

User applied rule all left

And obtained the following new goals

• Proof step 12:

  a = a ∨ a ∈ ax0 ↔ a ∈ ay, # ax0 + 1 ≠ # ay, unique_al(ax0), unique_al(ay), ¬ a ∈ ax0, ∀ a0. a0 = a ∨ a0 ∈ ax0 ↔ a0 ∈ ay, ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0) ⊦

Rule argument

Additional information

Used terms: a

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  # ax0 + 1 ≠ # ay, unique_al(ay), unique_al(ax0), a ∈ ay, ¬ a ∈ ax0, ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0), ∀ a0. a0 = a ∨ a0 ∈ ax0 ↔ a0 ∈ ay ⊦

Rule argument

Additional information

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 14:

  ⊦ a ∈ ax ↔ (∃ d, ay, az. ax = ay + (a × d)' + az)

• Proof step 15:

  # ax0 + 1 ≠ # ay, unique_al(ax0), unique_al(ay), a ∈ ay, ¬ a ∈ ax0, a ∈ ay ↔ (∃ d, ay0, az. ay = ay0 + (a × d)' + az), ∀ a0. a0 = a ∨ a0 ∈ ax0 ↔ a0 ∈ ay, ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0) ⊦

Rule argument

Additional information

Used terms: a, ay

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

Lemma ex

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  # ax0 ≠ # ay0 + # az, unique_al(ax0), unique_al(ay0 + (a × d)' + az), ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0), ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az) ⊦

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ a ∈ ax ∨ a ∈ ay; inone: ⊦ a ∈ (b × d0)' ↔ a = b; nuni: ⊦ unique_al(ay + (a × d)' + az) → ¬ a ∈ ay; nuni-01: ⊦ unique_al(ay + (a × d)' + az) → ¬ a ∈ az; a: ⊦ (x + y) + z = x + y + z; from specification list[assoclist] c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n = m + n0 ↔ n = n0; (s) ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s) ⊦ #(x + y) = # x + # y; (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 17:

  # ax0 ≠ # ay0 + # az, unique_al(ay0 + (a × d)' + az), unique_al(ax0), ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az), ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0), # ax0 = #(ay0 + az) ∨ ¬ unique_al(ay0 + az) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0 + az) ⊦

Rule argument

Additional information

Used terms: ay0 + az

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  # ax0 ≠ # ay0 + # az, ¬(a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az), unique_al(ax0), unique_al(ay0 + (a × d)' + az), ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0), ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az) ⊦

Rule argument

Additional information

Used simplifier rules: app: ⊦ unique_al(ax + p ' + ay) → unique_al(ax + ay); inapp: ⊦ a ∈ ax + ay ↔ a ∈ ax ∨ a ∈ ay; a: ⊦ (x + y) + z = x + y + z; from specification list[assoclist] ⊦ #(x + y) = # x + # y; (s)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

• Proof step 19:

  # ax0 ≠ # ay0 + # az, ¬(a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az), unique_al(ay0 + (a × d)' + az), unique_al(ax0), a0 ∈ ax0, ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az), ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0) ⊦

• Proof step 21:

  # ax0 ≠ # ay0 + # az, ¬(a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az), unique_al(ay0 + (a × d)' + az), unique_al(ax0), ¬ a0 ∈ ax0, ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az), ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0) ⊦

Rule argument

Additional information

Used formulas: a0 ∈ ax0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  # ax0 ≠ # ay0 + # az, unique_al(ax0), unique_al(ay0 + (a × d)' + az), a0 ∈ ax0, ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ¬ a0 ∈ ay0, ¬ a0 ∈ az, ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0), ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az) ⊦

Rule argument

Additional information

Proof step 20 from proof uniquelen (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: a0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  # ax0 ≠ # ay0 + # az, unique_al(ax0), unique_al(ay0 + (a × d)' + az), ¬ a0 ∈ ax0, ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0), ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az), a0 ∈ ay0 ∨ a0 ∈ az ⊦

Rule argument

Additional information

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

Heuristic cut applied rule cut formula

And obtained the following new goals

• Proof step 23:

  # ax0 ≠ # ay0 + # az, unique_al(ay0 + (a × d)' + az), unique_al(ax0), a0 ∈ az, ¬ a0 ∈ ax0, ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az), ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0), a0 ∈ ay0 ∨ a0 ∈ az ⊦

• Proof step 25:

  # ax0 ≠ # ay0 + # az, unique_al(ay0 + (a × d)' + az), unique_al(ax0), ¬ a0 ∈ az, ¬ a0 ∈ ax0, ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az), ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0), a0 ∈ ay0 ∨ a0 ∈ az ⊦

Rule argument

Additional information

Used formulas: a0 ∈ az

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  # ax0 ≠ # ay0 + # az, unique_al(ax0), unique_al(ay0 + (a × d)' + az), a0 ∈ az, ¬ a0 ∈ ax0, ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0), ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az) ⊦

Rule argument

Additional information

Proof step 24 from proof uniquelen (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: a0

Proof step 25 from proof uniquelen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  # ax0 ≠ # ay0 + # az, unique_al(ax0), unique_al(ay0 + (a × d)' + az), a0 ∈ ay0, ¬ a0 ∈ az, ¬ a0 ∈ ax0, ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ∀ ay0. # ax0 = # ay0 ∨ ¬ unique_al(ay0) ∨ ¬(∀ a. a ∈ ax0 ↔ a ∈ ay0), ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az) ⊦

Rule argument

Additional information

Proof step 26 from proof uniquelen (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: a0