Proof step 1 from proof lenequal (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), b ∈ ay, ¬ b ∈ ax, ∀ a. a ∈ ax → a ∈ ay ⊦

Rule argument

Additional information

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

User applied rule structural induction

And obtained the following new goals

• Proof step 3:

  #(pair + ax0) = # ay, unique_al(pair + ax0), unique_al(ay), b ∈ ay, ¬ b ∈ pair + ax0, ∀ a. a ∈ pair + ax0 → a ∈ ay, ∀ ay, b. ¬(# ax0 = # ay ∧ unique_al(ay) ∧ unique_al(ax0) ∧ b ∈ ay ∧ ¬ b ∈ ax0 ∧ (∀ a. a ∈ ax0 → a ∈ ay)) ⊦

Rule argument

Additional information

Used simplifier rules: assoclist freely generated by [], +;; ⊦ # [] = 0; (s) ⊦ # x = 0 ↔ x = []; (s)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 4:

  # ax0 + 1 = # ay, unique_al(ay), unique_al(pair ' + ax0), b ∈ ay, ¬ b ∈ pair ', ¬ b ∈ ax0, ∀ ay0, b0. # ax0 ≠ # ay0 ∨ ¬ unique_al(ay0) ∨ ¬ unique_al(ax0) ∨ ¬ b0 ∈ ay0 ∨ b0 ∈ 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 4 from proof lenequal (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 5:

  # ax0 + 1 = # ay, unique_al((a × d)' + ax0), unique_al(ay), b ∈ ay, ¬ b ∈ (a × d)', ¬ b ∈ ax0, ∀ a0. a0 ∈ (a × d)' ∨ a0 ∈ ax0 → a0 ∈ ay, ∀ ay0, b0. # ax0 ≠ # ay0 ∨ ¬ unique_al(ay0) ∨ ¬ unique_al(ax0) ∨ ¬ b0 ∈ ay0 ∨ b0 ∈ ax0 ∨ ¬(∀ a. a ∈ ax0 → a ∈ ay0) ⊦

Rule argument

Additional information

Used simplifier rules: pair freely generated by ×;;

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  # ax0 + 1 = # ay, b ≠ a, unique_al(ay), unique_al(ax0), b ∈ ay, ¬ b ∈ ax0, ¬ a ∈ ax0, ∀ ay0, b0. # ax0 ≠ # ay0 ∨ ¬ unique_al(ay0) ∨ ¬ b0 ∈ ay0 ∨ b0 ∈ ax0 ∨ ¬(∀ 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 6 from proof lenequal (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 7:

  # ax0 + 1 = # ay, b ≠ a, unique_al(ax0), unique_al(ay), b ∈ ay, ¬ b ∈ ax0, ¬ a ∈ ax0, a = a ∨ a ∈ ax0 → a ∈ ay, ∀ a0. a0 ∈ ax0 → a0 ∈ ay, ∀ ay0, b0. # ax0 ≠ # ay0 ∨ ¬ unique_al(ay0) ∨ ¬ b0 ∈ ay0 ∨ b0 ∈ ax0 ∨ ¬(∀ a. a ∈ ax0 → a ∈ ay0) ⊦

Rule argument

Additional information

Used terms: a

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  # ax0 + 1 = # ay, b ≠ a, unique_al(ay), unique_al(ax0), b ∈ ay, a ∈ ay, ¬ b ∈ ax0, ¬ a ∈ ax0, ∀ ay0, b0. # ax0 ≠ # ay0 ∨ ¬ unique_al(ay0) ∨ ¬ b0 ∈ ay0 ∨ b0 ∈ ax0 ∨ ¬(∀ a. a ∈ ax0 → a ∈ ay0), ∀ a0. a0 ∈ ax0 → a0 ∈ ay ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

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

• Proof step 10:

  # ax0 + 1 = # ay, b ≠ a, unique_al(ax0), unique_al(ay), b ∈ ay, ¬ b ∈ ax0, ¬ a ∈ ax0, ∀ a0. a0 ∈ ax0 → a0 ∈ ay, ∀ ay0, b0. # ax0 ≠ # ay0 ∨ ¬ unique_al(ay0) ∨ ¬ b0 ∈ ay0 ∨ b0 ∈ ax0 ∨ ¬(∀ a. a ∈ ax0 → a ∈ ay0), ∃ d, ay0, az. ay = ay0 + (a × d)' + az ⊦

Rule argument

Additional information

Used terms: ay, a

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

Lemma ex

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  # ax0 = # ay0 + # az, b ≠ a, unique_al(ay0 + (a × d)' + az), unique_al(ax0), ¬ b ∈ ax0, ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ∀ ay0, b0. # ax0 ≠ # ay0 ∨ ¬ unique_al(ay0) ∨ ¬ b0 ∈ ay0 ∨ b0 ∈ ax0 ∨ ¬(∀ a. a ∈ ax0 → a ∈ ay0), ∀ a0. a0 ∈ ax0 → a0 ∈ ay0 ∨ a0 ∈ az, b ∈ ay0 ∨ b ∈ az ⊦

Rule argument

Additional information

Used simplifier rules: inkeys-n: ⊦ a ≠ b → (a ∈ (b × d0)' + ax ↔ a ∈ ax); 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 11 from proof lenequal (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 12:

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

Rule argument

Additional information

Used terms: ay0 + az, b

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  # ax0 = # ay0 + # az, b ≠ a, unique_al(ay0 + (a × d)' + az), unique_al(ax0), a0 ∈ ax0, ¬ b ∈ ax0, ¬ a ∈ ax0, ¬ a ∈ ay0, ¬ a ∈ az, ¬ a0 ∈ ay0, ¬ a0 ∈ az, ∀ a0. a0 ∈ ax0 → a0 ∈ ay0 ∨ a0 ∈ az, b ∈ ay0 ∨ b ∈ 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 13 from proof lenequal (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: a0