Proof step 1 from proof inapp-06 (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(x + a ' + y = x0 + a0 ' + z ↔ x = x0 ∧ a = a0 ∧ y = z), ¬ a ∈ z, ¬ a0 ∈ y ⊦

Rule argument

Additional information

Proof step 2 from proof inapp-06 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  x + a ' + y = x0 + a0 ' + z, ¬ a ∈ z, ¬ a0 ∈ y ⊦ x = x0 ∧ a = a0 ∧ y = z

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z;

Proof step 3 from proof inapp-06 (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 4:

  (a1 + x1) + a ' + y = x0 + a0 ' + z, ¬ a ∈ z, ¬ a0 ∈ y, ∀ a, y, x0, a0, z. x1 + a ' + y = x0 + a0 ' + z ∧ ¬ a ∈ z ∧ ¬ a0 ∈ y → x1 = x0 ∧ a = a0 ∧ y = z ⊦ a1 + x1 = x0 ∧ a = a0 ∧ y = z

Rule argument

Additional information

Used simplifier rules: isin: ⊦ a ' + x = y + a0 ' + z ∧ y ≠ [] → a0 ∈ x; neq-02: ⊦ ¬ a ∈ z ∧ a ≠ a0 → a0 ' + z ≠ y1 + a ' + x; Nil: ⊦ [] + x = x; c-06: ⊦ a ' + x = a0 ' + y ↔ a = a0 ∧ x = y; a: ⊦ (x + y) + z = x + y + z; list('a) freely generated by [], +;;

Proof step 4 from proof inapp-06 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  (a1 ' + x1) + a ' + y = x0 + a0 ' + z, ¬ a ∈ z, ¬ a0 ∈ y, ∀ a, y, x0, a0, z. x1 + a ' + y = x0 + a0 ' + z ∧ ¬ a ∈ z ∧ ¬ a0 ∈ y → x1 = x0 ∧ a = a0 ∧ y = z ⊦ a1 ' + x1 = x0 ∧ a = a0 ∧ y = z

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; c: ⊦ a + x = a ' + x;

Proof step 5 from proof inapp-06 (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 6:

  (a1 ' + x1) + a ' + y = (a2 + x) + a0 ' + z, ¬ a ∈ z, ¬ a0 ∈ y, ∀ a, y, x0, a0, z. x1 + a ' + y = x0 + a0 ' + z ∧ ¬ a ∈ z ∧ ¬ a0 ∈ y → x1 = x0 ∧ a = a0 ∧ y = z, ∀ a1, x1, a, y, a0, z. (a1 ' + x1) + a ' + y = x + a0 ' + z ∧ ¬ a ∈ z ∧ ¬ a0 ∈ y ∧ (∀ a, y, x0, a0, z. x1 + a ' + y = x0 + a0 ' + z ∧ ¬ a ∈ z ∧ ¬ a0 ∈ y → x1 = x0 ∧ a = a0 ∧ y = z) → a1 ' + x1 = x ∧ a = a0 ∧ y = z ⊦ a1 ' + x1 = a2 + x ∧ a = a0 ∧ y = z

Rule argument

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x; c-06: ⊦ a ' + x = a0 ' + y ↔ a = a0 ∧ x = y; neq: ⊦ ¬ a ∈ x → x ≠ y + a ' + z; a: ⊦ (x + y) + z = x + y + z; list('a) freely generated by [], +;;

Proof step 6 from proof inapp-06 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  x1 + a ' + y = x + a0 ' + z, ¬ a0 ∈ y, ¬ a ∈ z, ∀ a, y, x0, a0, z. x1 + a ' + y = x0 + a0 ' + z ∧ ¬ a ∈ z ∧ ¬ a0 ∈ y → x1 = x0 ∧ a = a0 ∧ y = z, ∀ a1, x1, a, y, a0, z. (a1 ' + x1) + a ' + y = x + a0 ' + z ∧ ¬ a ∈ z ∧ ¬ a0 ∈ y ∧ (∀ a, y, x0, a0, z. x1 + a ' + y = x0 + a0 ' + z ∧ ¬ a ∈ z ∧ ¬ a0 ∈ y → x1 = x0 ∧ a = a0 ∧ y = z) → a1 ' + x1 = x ∧ a = a0 ∧ y = z ⊦ x1 = x ∧ a = a0 ∧ y = z

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x; c-06: ⊦ a ' + x = a0 ' + y ↔ a = a0 ∧ x = y; a: ⊦ (x + y) + z = x + y + z;

Proof step 7 from proof inapp-06 (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: a, y, x, a0, z