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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(a ∈ ax + ax0 ↔ a ∈ ax ∨ a ∈ ax0) ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ¬(a ∈ ax + ax0 ↔ a ∈ ax ∨ a ∈ ax0) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ a ∈ ax ↔ (∃ d. (a, d) ∈ ax)

• Proof step 5:

  ⊦ (∃ d. (a, d) ∈ ax + ax0) ↔ a ∈ ax ∨ a ∈ ax0

Rule argument

Additional information

Used terms: ax + ax0, a

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

Lemma inassoc

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  ⊦ (∃ d. (a, d) ∈ ax ∨ (a, d) ∈ ax0) ↔ a ∈ ax ∨ a ∈ ax0

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ a ∈ ax ↔ (∃ d. (a, d) ∈ ax)

• Proof step 8:

  ⊦ (∃ d. (a, d) ∈ ax ∨ (a, d) ∈ ax0) ↔ (∃ d. (a, d) ∈ ax) ∨ a ∈ ax0

Rule argument

Additional information

Used terms: ax, a

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

Lemma inassoc

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

  ⊦ a ∈ ax ↔ (∃ d. (a, d) ∈ ax)

• Proof step 10:

  ⊦ (∃ d. (a, d) ∈ ax ∨ (a, d) ∈ ax0) ↔ (∃ d. (a, d) ∈ ax) ∨ (∃ d. (a, d) ∈ ax0)

Rule argument

Additional information

Used terms: ax0, a

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

Lemma inassoc

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

User applied rule case distinction

And obtained the following new goals

• Proof step 11:

  ∃ d. (a, d) ∈ ax ∨ (a, d) ∈ ax0 ⊦ ∃ d. (a, d) ∈ ax, ∃ d. (a, d) ∈ ax0

• Proof step 15:

  (∃ d. (a, d) ∈ ax) ∨ (∃ d. (a, d) ∈ ax0) ⊦ ∃ d. (a, d) ∈ ax ∨ (a, d) ∈ ax0

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  (a, d) ∈ ax ∨ (a, d) ∈ ax0 ⊦ ∃ d. (a, d) ∈ ax0, ∃ d. (a, d) ∈ ax

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 13:

  (a, d) ∈ ax ⊦ ∃ d. (a, d) ∈ ax, ∃ d. (a, d) ∈ ax0

• Proof step 14:

  (a, d) ∈ ax0 ⊦ ∃ d. (a, d) ∈ ax, ∃ d. (a, d) ∈ ax0

Rule argument

Additional information

Proof step 13 from proof inapp (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 14 from proof inapp (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 15 from proof inapp (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

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

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 17:

  (a, d) ∈ ax ⊦ ∃ d. (a, d) ∈ ax ∨ (a, d) ∈ ax0

• Proof step 18:

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

Rule argument

Additional information

Proof step 17 from proof inapp (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 18 from proof inapp (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used terms: d0