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

Heuristic structural induction applied rule structural induction

And obtained the following new goals

• Proof step 2:

  ∀ a. a ∈ ax0 ↔ (∃ d. (a, d) ∈ ax0) ⊦ a ∈ p + ax0 ↔ (∃ d. (a, d) ∈ p + ax0)

Rule argument

Additional information

Used simplifier rules: empty: ⊦ ¬ a ∈ []; List[A] freely generated by [], +; ⊦ ¬ a ∈ []; (s)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ∀ a. a ∈ ax0 ↔ (∃ d. (a, d) ∈ ax0) ⊦ a ∈ p ' + ax0 ↔ (∃ d. (a, d) = p ∨ (a, d) ∈ ax0)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ a0 ' + x ↔ a = a0 ∨ a ∈ x; (s) ⊦ a + x = a ' + x; (s)

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

Heuristic expand tuples applied rule constructor cut

And obtained the following new goals

• Proof step 4:

  p = (a0, d0), ∀ a. a ∈ ax0 ↔ (∃ d. (a, d) ∈ ax0) ⊦ a ∈ p ' + ax0 ↔ (∃ d. (a, d) = p ∨ (a, d) ∈ ax0)

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  ∀ a. a ∈ ax0 ↔ (∃ d. (a, d) ∈ ax0) ⊦ a = a0 ∨ a ∈ ax0 ↔ (∃ d. (a, d) = (a0, d0) ∨ (a, d) ∈ ax0)

Rule argument

Additional information

Used simplifier rules: inrec: ⊦ a ∈ (b, d0) ' + ax ↔ a = b ∨ a ∈ ax; a: ⊦ (x + y) + z = x + y + z; from specification list

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

• Proof step 6:

  ∀ a. a ∈ ax0 ↔ (∃ d. (a, d) ∈ ax0), a = a0 ∨ a ∈ ax0 ⊦ ∃ d. (a, d) = (a0, d0) ∨ (a, d) ∈ ax0

• Proof step 12:

  a ≠ a0, ¬ a ∈ ax0, ∀ a. a ∈ ax0 ↔ (∃ d. (a, d) ∈ ax0), ∃ d. (a, d) = (a0, d0) ∨ (a, d) ∈ ax0 ⊦

Rule argument

Additional information

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

• Proof step 7:

  a = a0, ∀ a. a ∈ ax0 ↔ (∃ d. (a, d) ∈ ax0) ⊦ ∃ d. (a, d) = (a0, d0) ∨ (a, d) ∈ ax0

• Proof step 9:

  a ∈ ax0, ∀ a. a ∈ ax0 ↔ (∃ d. (a, d) ∈ ax0) ⊦ ∃ d. (a, d) = (a0, d0) ∨ (a, d) ∈ ax0

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  ∀ a. a ∈ ax0 ↔ (∃ d. (a, d) ∈ ax0) ⊦ ∃ d. d = d0 ∨ (a0, d) ∈ ax0

Rule argument

Additional information

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used terms: d0

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

User applied rule all left

And obtained the following new goals

• Proof step 10:

  a ∈ ax0, a ∈ ax0 ↔ (∃ d. (a, d) ∈ ax0) ⊦ ∃ d. (a, d) = (a0, d0) ∨ (a, d) ∈ ax0

Rule argument

Additional information

Used terms: a

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  a ∈ ax0, (a, d) ∈ ax0 ⊦ ∃ d. (a, d) = (a0, d0) ∨ (a, d) ∈ ax0

Rule argument

Additional information

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used terms: d

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  (a, d) ≠ (a0, d0), a ≠ a0, (a, d) ∈ ax0, ¬ a ∈ ax0, ∀ a. a ∈ ax0 ↔ (∃ d. (a, d) ∈ ax0) ⊦

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 14:

  (a, d) ≠ (a0, d0), a ≠ a0, (a, d) ∈ ax0, ¬ a ∈ ax0, a ∈ ax0 ↔ (∃ d. (a, d) ∈ ax0) ⊦

Rule argument

Additional information

Used terms: a

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  (a, d) ≠ (a0, d0), a ≠ a0, (a, d) ∈ ax0, ¬ a ∈ ax0 ⊦ ∃ d. (a, d) ∈ ax0

Rule argument

Additional information

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used terms: d