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

⊦ a ∈ x → (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y)

System applied rule normalize

And obtained the following new goals

Proof step 2:
a ∈ x ⊦ ∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y

Rule argument

Additional information

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

a ∈ x ⊦ ∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ x ⊦ ∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y

Rule argument

Additional information

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

a ∈ x ⊦ ∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y

User applied rule structural induction

And obtained the following new goals

Proof step 4:
a ∈ a0 + x0, ∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y) ⊦ ∃ y, z. a0 + x0 = y + a ' + z ∧ ¬ a ∈ y

Rule argument

indvar: x

Additional information

Used simplifier rules: in-empty: ⊦ ¬ a ∈ []; list freely generated by [], +;;

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y), a = a0 ∨ a ∈ x0 ⊦ ∃ y, z. a0 ' + x0 = y + a ' + z ∧ ¬ a ∈ y

Rule argument

Additional information

Used simplifier rules: in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; in-17: ⊦ a ∈ b ' ↔ a = b; a: ⊦ (x + y) + z = x + y + z; c: ⊦ a + x = a ' + x;

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

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

User applied rule all left

And obtained the following new goals

Proof step 6:
a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y), ∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y), a = a0 ∨ a ∈ x0 ⊦ ∃ y, z. a0 ' + x0 = y + a ' + z ∧ ¬ a ∈ y

Rule argument

pos: left 1 terms: a abort: F discard: F computed: T

Additional information

Used terms: a

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a ∈ x0 → x0 = y + a ' + z ∧ ¬ a ∈ y, ∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y), a = a0 ∨ a ∈ x0 ⊦ ∃ y, z. a0 ' + x0 = y + a ' + z ∧ ¬ a ∈ y

Rule argument

Additional information

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

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

User applied rule case distinction

And obtained the following new goals

Proof step 8:
¬ a ∈ x0, ∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y), a = a0 ∨ a ∈ x0 ⊦ ∃ y, z. a0 ' + x0 = y + a ' + z ∧ ¬ a ∈ y
Proof step 10:
x0 = y + a ' + z, ¬ a ∈ y, ∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y), a = a0 ∨ a ∈ x0 ⊦ ∃ y, z. a0 ' + x0 = y + a ' + z ∧ ¬ a ∈ y

Rule argument

pos: left 1

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
¬ a0 ∈ x0, ∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y) ⊦ ∃ y, z. a0 ' + x0 = y + a0 ' + z ∧ ¬ a0 ∈ y

Rule argument

Additional information

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

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], x0

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x; app-07: ⊦ a ' + x = b ' + x ↔ a = b; a: ⊦ (x + y) + z = x + y + z; in-empty: ⊦ ¬ a ∈ []; Used terms: [], x0

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
¬ a ∈ y, ∀ a1. a1 ∈ y ∨ a1 ∈ a ' ∨ a1 ∈ z → (∃ y0, z0. y + a ' + z = y0 + a1 ' + z0 ∧ ¬ a1 ∈ y0), a = a0 ∨ a ∈ a ' ∨ a ∈ z ⊦ ∃ y0, z0. a0 ' + y + a ' + z = y0 + a ' + z0 ∧ ¬ a ∈ y0

Rule argument

Additional information

Used simplifier rules: in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; a: ⊦ (x + y) + z = x + y + z;

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

¬ a ∈ y, ∀ a1. a1 ∈ y ∨ a1 ∈ a ' ∨ a1 ∈ z → (∃ y0, z0. y + a ' + z = y0 + a1 ' + z0 ∧ ¬ a1 ∈ y0), a = a0 ∨ a ∈ a ' ∨ a ∈ z ⊦ ∃ y0, z0. a0 ' + y + a ' + z = y0 + a ' + z0 ∧ ¬ a ∈ y0

User applied rule exists right

And obtained the following new goals

Proof step 12:
¬ a ∈ y, ∀ a1. a1 ∈ y ∨ a1 ∈ a ' ∨ a1 ∈ z → (∃ y0, z0. y + a ' + z = y0 + a1 ' + z0 ∧ ¬ a1 ∈ y0), a = a0 ∨ a ∈ a ' ∨ a ∈ z ⊦ ∃ y0, z0. a0 ' + y + a ' + z = y0 + a ' + z0 ∧ ¬ a ∈ y0, a0 ' + y + a ' + z = (a0 ' + y) + a ' + z ∧ ¬ a ∈ a0 ' + y

Rule argument

pos: right 1 terms: a0 ' + y, z abort: F discard: F computed: T

Additional information

Used terms: a0 ' + y, z

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

¬ a ∈ y, ∀ a1. a1 ∈ y ∨ a1 ∈ a ' ∨ a1 ∈ z → (∃ y0, z0. y + a ' + z = y0 + a1 ' + z0 ∧ ¬ a1 ∈ y0), a = a0 ∨ a ∈ a ' ∨ a ∈ z ⊦ ∃ y0, z0. a0 ' + y + a ' + z = y0 + a ' + z0 ∧ ¬ a ∈ y0, a0 ' + y + a ' + z = (a0 ' + y) + a ' + z ∧ ¬ a ∈ a0 ' + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
¬ a0 ∈ y, ∀ a1. a1 ∈ y ∨ a1 ∈ a0 ' ∨ a1 ∈ z → (∃ y0, z0. y + a0 ' + z = y0 + a1 ' + z0 ∧ ¬ a1 ∈ y0) ⊦ ∃ y0, z0. a0 ' + y + a0 ' + z = y0 + a0 ' + z0 ∧ ¬ a0 ∈ y0

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ a = b ∨ a ∈ x; a: ⊦ (x + y) + z = x + y + z;

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

¬ a0 ∈ y, ∀ a1. a1 ∈ y ∨ a1 ∈ a0 ' ∨ a1 ∈ z → (∃ y0, z0. y + a0 ' + z = y0 + a1 ' + z0 ∧ ¬ a1 ∈ y0) ⊦ ∃ y0, z0. a0 ' + y + a0 ' + z = y0 + a0 ' + z0 ∧ ¬ a0 ∈ y0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], y + a0 ' + z abort: F discard: F computed: F

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x; app-08: ⊦ a ' + x = a ' + y ↔ x = y; a: ⊦ (x + y) + z = x + y + z; in-empty: ⊦ ¬ a ∈ []; Used terms: [], y + a0 ' + z

Proof Tree for firstoc (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.