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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  a ∈ ax ⊦ ∃ ax0, d, ax1. ax = ax0 + (a, d) ' + ax1 ∧ ¬ a ∈ ax0

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  a ∈ ax ⊦ ∃ ax0, d, ax1. ax = ax0 + (a, d) ' + ax1 ∧ ¬ a ∈ ax0

Rule argument

Additional information

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

User applied rule structural induction

And obtained the following new goals

• Proof step 4:

  a ∈ p + ax0, ∀ a. a ∈ ax0 → (∃ ax1, d, ax2. ax0 = ax1 + (a, d) ' + ax2 ∧ ¬ a ∈ ax1) ⊦ ∃ ax1, d, ax2. p + ax0 = ax1 + (a, d) ' + ax2 ∧ ¬ a ∈ ax1

Rule argument

Additional information

Used simplifier rules: empty: ⊦ ¬ a ∈ []; list('a) freely generated by [], +;;

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  ∀ a. a ∈ ax0 → (∃ ax1, d, ax2. ax0 = ax1 + (a, d) ' + ax2 ∧ ¬ a ∈ ax1), a ∈ p ' ∨ a ∈ ax0 ⊦ ∃ ax1, d, ax2. p ' + ax0 = ax1 + (a, d) ' + ax2 ∧ ¬ a ∈ ax1

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ax0 ↔ a ∈ ax ∨ a ∈ ax0; ⊦ a + x = a ' + x; (s) a: ⊦ (x + y) + z = x + y + z; from specification polylist

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

User applied rule structural induction

And obtained the following new goals

• Proof step 6:

  ∀ a. a ∈ ax0 → (∃ ax1, d, ax2. ax0 = ax1 + (a, d) ' + ax2 ∧ ¬ a ∈ ax1), a ∈ (a0, d) ' ∨ a ∈ ax0 ⊦ ∃ ax1, d0, ax2. (a0, d) ' + ax0 = ax1 + (a, d0) ' + ax2 ∧ ¬ a ∈ ax1

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  ∀ a. a ∈ ax0 → (∃ ax1, d, ax2. ax0 = ax1 + (a, d) ' + ax2 ∧ ¬ a ∈ ax1), a = a0 ∨ a ∈ ax0 ⊦ ∃ ax1, d0, ax2. (a0, d) ' + ax0 = ax1 + (a, d0) ' + ax2 ∧ ¬ a ∈ ax1

Rule argument

Additional information

Used simplifier rules: inone: ⊦ a ∈ (b, d0) ' ↔ a = b;

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

User applied rule cut formula

And obtained the following new goals

• Proof step 8:

  a = a0, ∀ a. a ∈ ax0 → (∃ ax1, d, ax2. ax0 = ax1 + (a, d) ' + ax2 ∧ ¬ a ∈ ax1), a = a0 ∨ a ∈ ax0 ⊦ ∃ ax1, d0, ax2. (a0, d) ' + ax0 = ax1 + (a, d0) ' + ax2 ∧ ¬ a ∈ ax1

• Proof step 10:

  a ≠ a0, ∀ a. a ∈ ax0 → (∃ ax1, d, ax2. ax0 = ax1 + (a, d) ' + ax2 ∧ ¬ a ∈ ax1), a = a0 ∨ a ∈ ax0 ⊦ ∃ ax1, d0, ax2. (a0, d) ' + ax0 = ax1 + (a, d0) ' + ax2 ∧ ¬ a ∈ ax1

Rule argument

Additional information

Used formulas: a = a0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  ∀ a. a ∈ ax0 → (∃ ax1, d, ax2. ax0 = ax1 + (a, d) ' + ax2 ∧ ¬ a ∈ ax1) ⊦ ∃ ax1, d0, ax2. (a0, d) ' + ax0 = ax1 + (a0, d0) ' + ax2 ∧ ¬ a0 ∈ ax1

Rule argument

Additional information

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: empty: ⊦ ¬ a ∈ []; ⊦ [] + x = x; (s) ⊦ # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0); (ws) a: ⊦ (x + y) + z = x + y + z; from specification polylist Used terms: [], d, ax0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  a ≠ a0, a ∈ ax0, ∀ a. a ∈ ax0 → (∃ ax1, d, ax2. ax0 = ax1 + (a, d) ' + ax2 ∧ ¬ a ∈ ax1) ⊦ ∃ ax1, d0, ax2. (a0, d) ' + ax0 = ax1 + (a, d0) ' + ax2 ∧ ¬ a ∈ ax1

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 12:

  a ≠ a0, a ∈ ax0, a ∈ ax0 → (∃ ax1, d, ax2. ax0 = ax1 + (a, d) ' + ax2 ∧ ¬ a ∈ ax1) ⊦ ∃ ax1, d0, ax2. (a0, d) ' + ax0 = ax1 + (a, d0) ' + ax2 ∧ ¬ a ∈ ax1

Rule argument

Additional information

Used terms: a

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  a ≠ a0, ¬ a ∈ ax1 ⊦ ∃ ax, d1, ax3. (a0, d) ' + ax1 + (a, d0) ' + ax2 = ax + (a, d1) ' + ax3 ∧ ¬ a ∈ ax

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ax0 ↔ a ∈ ax ∨ a ∈ ax0; inone: ⊦ a ∈ (b, d0) ' ↔ a = b; a: ⊦ (x + y) + z = x + y + z; from specification polylist

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: inkeys-n: ⊦ a ≠ b → (a ∈ (b, d0) ' + ax ↔ a ∈ ax); a: ⊦ (x + y) + z = x + y + z; from specification polylist Used terms: (a0, d) ' + ax1, d0, ax2