Proof step 1 from proof ordered-ex-divide-lels (conclusion) (back to tree)

ordered≤(x) ⊦ ∃ y, z. x = y + z ∧ (∀ b. b ∈ y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)

User applied rule structural induction

And obtained the following new goals

Proof step 2:
ordered≤([]) ⊦ ∃ y, z. [] = y + z ∧ (∀ b. b ∈ y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)
Proof step 4:
ordered≤(a0 + x0), ∀ a. ordered≤(x0) → (∃ y, z. x0 = y + z ∧ (∀ b. b ∈ y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)) ⊦ ∃ y, z. a0 + x0 = y + z ∧ (∀ b. b ∈ y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)

Rule argument

indvar: x

Additional information

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

Proof step 2 from proof ordered-ex-divide-lels (previous step) (back to tree)

ordered≤([]) ⊦ ∃ y, z. [] = y + z ∧ (∀ b. b ∈ y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ ∃ y, z. [] = y + z ∧ (∀ b. b ∈ y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)

Rule argument

Additional information

Used simplifier rules: le-e: ⊦ ordered≤([]);

Proof step 3 from proof ordered-ex-divide-lels (previous step) (back to tree)

⊦ ∃ y, z. [] = y + z ∧ (∀ b. b ∈ y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], [] abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ [] + x = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ ¬ a ∈ []; (s) Used terms: [], []

Proof step 4 from proof ordered-ex-divide-lels (previous step) (back to tree)

ordered≤(a0 + x0), ∀ a. ordered≤(x0) → (∃ y, z. x0 = y + z ∧ (∀ b. b ∈ y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)) ⊦ ∃ y, z. a0 + x0 = y + z ∧ (∀ b. b ∈ y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
ordered≤(a0 ' + x0), ∀ a1. ∃ y, z. x0 = y + z ∧ (∀ b. b ∈ y → ¬ a1 < b) ∧ (∀ b. b ∈ z → a1 < b) ⊦ ∃ y, z. a0 ' + x0 = y + z ∧ (∀ b. b ∈ y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)

Rule argument

Additional information

Used simplifier rules: cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x); app-01: ⊦ ordered≤(x + y) → ordered≤(y); a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a + x = a ' + x; (s)

Proof step 5 from proof ordered-ex-divide-lels (previous step) (back to tree)

ordered≤(a0 ' + x0), ∀ a1. ∃ y, z. x0 = y + z ∧ (∀ b. b ∈ y → ¬ a1 < b) ∧ (∀ b. b ∈ z → a1 < b) ⊦ ∃ y, z. a0 ' + x0 = y + z ∧ (∀ b. b ∈ y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)

User applied rule all left

And obtained the following new goals

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

Rule argument

pos: left 2 terms: a abort: F discard: T computed: F

Additional information

Used terms: a

Proof step 6 from proof ordered-ex-divide-lels (previous step) (back to tree)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
ordered≤(a0 ' + y + z), ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → ¬ a < b ⊦ ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → ¬ a < b) ∧ (∀ b. b ∈ z0 → a < b)

Rule argument

Additional information

Used simplifier rules: cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x); app-01: ⊦ ordered≤(x + y) → ordered≤(y); a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 7 from proof ordered-ex-divide-lels (previous step) (back to tree)

ordered≤(a0 ' + y + z), ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → ¬ a < b ⊦ ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → ¬ a < b) ∧ (∀ b. b ∈ z0 → a < b)

User applied rule exists right

And obtained the following new goals

Proof step 8:
ordered≤(a0 ' + y + z), ∀ b. b ∈ y → ¬ a < b, ∀ b. b ∈ z → a < b ⊦ ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → ¬ a < b) ∧ (∀ b. b ∈ z0 → a < b), a0 ' + y + z = (a0 ' + y) + z ∧ (∀ b. b ∈ a0 ' + y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)

Rule argument

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

Additional information

Used terms: a0 ' + y, z

Proof step 8 from proof ordered-ex-divide-lels (previous step) (back to tree)

ordered≤(a0 ' + y + z), ∀ b. b ∈ y → ¬ a < b, ∀ b. b ∈ z → a < b ⊦ ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → ¬ a < b) ∧ (∀ b. b ∈ z0 → a < b), a0 ' + y + z = (a0 ' + y) + z ∧ (∀ b. b ∈ a0 ' + y → ¬ a < b) ∧ (∀ b. b ∈ z → a < b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
a < b, ordered≤(a0 ' + y + z), ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → ¬ a < b, b = a0 ∨ b ∈ y ⊦ ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → ¬ a < b) ∧ (∀ b. b ∈ z0 → a < b)

Rule argument

Additional information

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

Proof step 9 from proof ordered-ex-divide-lels (previous step) (back to tree)

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

User applied rule cut formula

And obtained the following new goals

Proof step 10:
b = a0, a < b, ordered≤(a0 ' + y + z), ∀ b. b ∈ y → ¬ a < b, ∀ b. b ∈ z → a < b, b = a0 ∨ b ∈ y ⊦ ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → ¬ a < b) ∧ (∀ b. b ∈ z0 → a < b)
Proof step 19:
b ≠ a0, a < b, ordered≤(a0 ' + y + z), ∀ b. b ∈ y → ¬ a < b, ∀ b. b ∈ z → a < b, b = a0 ∨ b ∈ y ⊦ ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → ¬ a < b) ∧ (∀ b. b ∈ z0 → a < b)

Rule argument

fma: b = a0

Additional information

Used formulas: b = a0

Proof step 10 from proof ordered-ex-divide-lels (previous step) (back to tree)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Proof step 11 from proof ordered-ex-divide-lels (previous step) (back to tree)

a < a0, ordered≤(a0 ' + y + z), ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → ¬ a < b ⊦ ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → ¬ a < b) ∧ (∀ b. b ∈ z0 → a < b)

User applied rule exists right

And obtained the following new goals

Proof step 12:
a < a0, ordered≤(a0 ' + y + z), ∀ b. b ∈ y → ¬ a < b, ∀ b. b ∈ z → a < b ⊦ a0 ' + y + z = [] + a0 ' + y + z ∧ (∀ b. b ∈ [] → ¬ a < b) ∧ (∀ b. b ∈ a0 ' + y + z → a < b)

Rule argument

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

Additional information

Used terms: [], a0 ' + y + z

Proof step 12 from proof ordered-ex-divide-lels (previous step) (back to tree)

a < a0, ordered≤(a0 ' + y + z), ∀ b. b ∈ y → ¬ a < b, ∀ b. b ∈ z → a < b ⊦ a0 ' + y + z = [] + a0 ' + y + z ∧ (∀ b. b ∈ [] → ¬ a < b) ∧ (∀ b. b ∈ a0 ' + y + z → a < b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
a < a0, b < a0, ordered≤(a0 ' + y + z), ¬ a < b, ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → ¬ a < b, b ∈ y ∨ b ∈ z ⊦

Rule argument

Additional information

Used simplifier rules: nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification weakord a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s) ⊦ a ∈ a0 ' ↔ a = a0; (s) ⊦ ¬ a ∈ []; (s) ⊦ x + y = x + z ↔ y = z; (s) ⊦ [] + x = x; (s)

Proof step 13 from proof ordered-ex-divide-lels (previous step) (back to tree)

a < a0, b < a0, ordered≤(a0 ' + y + z), ¬ a < b, ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → ¬ a < b, b ∈ y ∨ b ∈ z ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 14:
b < a0, a < a0, ordered≤(a0 ' + y + z), b ∈ y, ¬ a < b, ∀ b. b ∈ y → ¬ a < b, ∀ b. b ∈ z → a < b ⊦
Proof step 17:
b < a0, a < a0, ordered≤(a0 ' + y + z), b ∈ z, ¬ a < b, ∀ b. b ∈ y → ¬ a < b, ∀ b. b ∈ z → a < b ⊦

Rule argument

pos: left 7

Additional information

Proof step 14 from proof ordered-ex-divide-lels (previous step) (back to tree)

b < a0, a < a0, ordered≤(a0 ' + y + z), b ∈ y, ¬ a < b, ∀ b. b ∈ y → ¬ a < b, ∀ b. b ∈ z → a < b ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
a < a0, b < a0, ordered≤(a0 ' + y + z), b ∈ y, ¬ a < b, ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → ¬ a < b ⊦

Rule argument

Additional information

Proof step 15 from proof ordered-ex-divide-lels (previous step) (back to tree)

a < a0, b < a0, ordered≤(a0 ' + y + z), b ∈ y, ¬ a < b, ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → ¬ a < b ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ b < a ∧ b ∈ x → ¬ ordered≤(a ' + x)

Rule argument

spec: <toplevel>, name:orderedle-ls, seq: ⊦ b < a ∧ b ∈ x → ¬ ordered≤(a ' + x), subst:[b, x, a] → [b, y + z, a0], <all paths>

Additional information

Used simplifier rules: ⊦ a ∈ x → a ∈ x + y; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: b, y + z, a0

Proof step 16 from proof ordered-ex-divide-lels (previous step) (back to tree)

⊦ b < a ∧ b ∈ x → ¬ ordered≤(a ' + x)

Lemma orderedle-ls

Proof step 17 from proof ordered-ex-divide-lels (previous step) (back to tree)

b < a0, a < a0, ordered≤(a0 ' + y + z), b ∈ z, ¬ a < b, ∀ b. b ∈ y → ¬ a < b, ∀ b. b ∈ z → a < b ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
a < a0, b < a0, ordered≤(a0 ' + y + z), b ∈ z, ¬ a < b, ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → ¬ a < b ⊦

Rule argument

Additional information

Proof step 18 from proof ordered-ex-divide-lels (previous step) (back to tree)

a < a0, b < a0, ordered≤(a0 ' + y + z), b ∈ z, ¬ a < b, ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → ¬ a < b ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: b

Additional information

Used terms: b

Proof step 19 from proof ordered-ex-divide-lels (previous step) (back to tree)

b ≠ a0, a < b, ordered≤(a0 ' + y + z), ∀ b. b ∈ y → ¬ a < b, ∀ b. b ∈ z → a < b, b = a0 ∨ b ∈ y ⊦ ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → ¬ a < b) ∧ (∀ b. b ∈ z0 → a < b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
b ≠ a0, a < b, ordered≤(a0 ' + y + z), b ∈ y, ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → ¬ a < b ⊦ ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → ¬ a < b) ∧ (∀ b. b ∈ z0 → a < b)

Rule argument

Additional information

Proof step 20 from proof ordered-ex-divide-lels (previous step) (back to tree)

b ≠ a0, a < b, ordered≤(a0 ' + y + z), b ∈ y, ∀ b. b ∈ z → a < b, ∀ b. b ∈ y → ¬ a < b ⊦ ∃ y0, z0. a0 ' + y + z = y0 + z0 ∧ (∀ b. b ∈ y0 → ¬ a < b) ∧ (∀ b. b ∈ z0 → a < b)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: b

Additional information

Used terms: b

Proof Tree for ordered-ex-divide-lels (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.