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

⊦ ins≤(a, x) = ins≤(a, y) ↔ x = y

System applied rule normalize

And obtained the following new goals

Proof step 2:
¬(ins≤(a, x) = ins≤(a, y) ↔ x = y) ⊦

Rule argument

Additional information

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

¬(ins≤(a, x) = ins≤(a, y) ↔ x = y) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
ins≤(a, x) = ins≤(a, y), x ≠ y ⊦

Rule argument

Additional information

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

ins≤(a, x) = ins≤(a, y), x ≠ y ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 4:
ins≤(a, []) = ins≤(a, y), [] ≠ y ⊦
Proof step 8:
ins≤(a, a0 + x0) = ins≤(a, y), a0 + x0 ≠ y, ∀ a, y. ¬(ins≤(a, x0) = ins≤(a, y) ∧ x0 ≠ y) ⊦

Rule argument

indvar: x

Additional information

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

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

ins≤(a, []) = ins≤(a, y), [] ≠ y ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ' = ins≤(a, y), [] ≠ y ⊦

Rule argument

Additional information

Used simplifier rules: ins-e: ⊦ ins≤(a, []) = a ';

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

a ' = ins≤(a, y), [] ≠ y ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 6:
a ' = ins≤(a, a0 + y0), [] ≠ a0 + y0, ∀ a. ¬(a ' = ins≤(a, y0) ∧ [] ≠ y0) ⊦

Rule argument

indvar: y

Additional information

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

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

a ' = ins≤(a, a0 + y0), [] ≠ a0 + y0, ∀ a. ¬(a ' = ins≤(a, y0) ∧ [] ≠ y0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a ' = ins≤(a, a0 ' + y0), (∀ a1. a1 ' ≠ ins≤(a1, y0)) ∨ [] = y0 ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ' + x ≠ []; (s) ⊦ a + x = a ' + x; (s)

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

a ' = ins≤(a, a0 ' + y0), (∀ a1. a1 ' ≠ ins≤(a1, y0)) ∨ [] = y0 ⊦

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: a0 < a

Additional information

Used simplifier rules: ins-n: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x); empty: ⊦ ins≤(a, x) ≠ []; ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; ⊦ a + x = a ' + x; (s) ⊦ a ' = a0 ' + x ↔ a = a0 ∧ x = []; (ws) ⊦ a ' + x ≠ []; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used formulas: a0 < a

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

ins≤(a, a0 + x0) = ins≤(a, y), a0 + x0 ≠ y, ∀ a, y. ¬(ins≤(a, x0) = ins≤(a, y) ∧ x0 ≠ y) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
ins≤(a, a0 ' + x0) = ins≤(a, y), a0 ' + x0 ≠ y, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)

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

ins≤(a, a0 ' + x0) = ins≤(a, y), a0 ' + x0 ≠ y, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 10:
ins≤(a, a0 ' + x0) = ins≤(a, y), a0 ' + x0 ≠ y, a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦
Proof step 19:
ins≤(a, a0 ' + x0) = ins≤(a, y), a0 ' + x0 ≠ y, ¬ a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

fma: a0 < a

Additional information

Used formulas: a0 < a

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

ins≤(a, a0 ' + x0) = ins≤(a, y), a0 ' + x0 ≠ y, a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
a0 ' + ins≤(a, x0) = ins≤(a, y), a0 ' + x0 ≠ y, a0 < a, a0 ∈ ins≤(a, y), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used simplifier rules: ins-n: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x); a: ⊦ (x + y) + z = x + y + z; from specification list in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list ⊦ a + x = a ' + x; (s)

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

a0 ' + ins≤(a, x0) = ins≤(a, y), a0 ' + x0 ≠ y, a0 < a, a0 ∈ ins≤(a, y), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ x ≠ [] ∧ x.head < a → ins≤(a, x) = x.head + ins≤(a, x.tail)
Proof step 13:
a0 ' + ins≤(a, x0) = ins≤(a, y), a0 ' + x0 ≠ y, a0 < a, a0 ∈ ins≤(a, y), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦ y ≠ [] ∧ y.head < a
Proof step 15:
a0 ' + ins≤(a, x0) = y.head + ins≤(a, y.tail), a0 ' + x0 ≠ y, y ≠ [], a0 < a, y.head < a, a0 ∈ y.head + ins≤(a, y.tail), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

spec: <toplevel>, name:ins-rec, seq: ⊦ x ≠ [] ∧ x.head < a → ins≤(a, x) = x.head + ins≤(a, x.tail), subst:[x, a] → [y, a], <all paths>

Additional information

Used terms: y, a

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

⊦ x ≠ [] ∧ x.head < a → ins≤(a, x) = x.head + ins≤(a, x.tail)

Lemma ins-rec

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

a0 ' + ins≤(a, x0) = ins≤(a, y), a0 ' + x0 ≠ y, a0 < a, a0 ∈ ins≤(a, y), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦ y ≠ [] ∧ y.head < a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
a0 ' + ins≤(a, x0) = ins≤(a, y), a0 ' + x0 ≠ y, y ≠ [], a0 < a, a0 ∈ ins≤(a, y), ¬ y.head < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used simplifier rules: ins-e: ⊦ ins≤(a, []) = a '; empty: ⊦ ins≤(a, x) ≠ []; a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ' = a0 ' + x ↔ a = a0 ∧ x = []; (ws)

Proof step 14 from proof eq (previous step) (back to tree)

a0 ' + ins≤(a, x0) = ins≤(a, y), a0 ' + x0 ≠ y, y ≠ [], a0 < a, a0 ∈ ins≤(a, y), ¬ y.head < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

spec: list-data, name:elim-add, seq: ⊦ x ≠ [] → (head = x.head ∧ tail = x.tail ↔ x = head + tail), subst:[x, head, tail] → [y, a2, y1]; ['a] → [tord]

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; ⊦ a ' + x = a0 ' + y ↔ a = a0 ∧ x = y; (ws) ⊦ a ' + x ≠ []; (s) ⊦ a + x = a ' + x; (s) in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list a: ⊦ (x + y) + z = x + y + z; from specification list lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification weakord[tord] ⊦ ¬ a < a; (s) ⊦ x ≠ [] → (head = x.head ∧ tail = x.tail ↔ x = head + tail); (elim) Used terms: y, a2, y1

Proof step 15 from proof eq (previous step) (back to tree)

a0 ' + ins≤(a, x0) = y.head + ins≤(a, y.tail), a0 ' + x0 ≠ y, y ≠ [], a0 < a, y.head < a, a0 ∈ y.head + ins≤(a, y.tail), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
ins≤(a, x0) = ins≤(a, y.tail), y.head ' + x0 ≠ y, y ≠ [], y.head < a, y.head ∈ y.head ' + ins≤(a, x0), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

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

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

ins≤(a, x0) = ins≤(a, y.tail), y.head ' + x0 ≠ y, y ≠ [], y.head < a, y.head ∈ y.head ' + ins≤(a, x0), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Proof step 17:
ins≤(a, x0) = ins≤(a, y1), y = a0 + y1, a0 ' + x0 ≠ y, y ≠ [], a0 < a, a0 ∈ a0 ' + ins≤(a, x0), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

spec: list-data, name:elim-add, seq: ⊦ x ≠ [] → (head = x.head ∧ tail = x.tail ↔ x = head + tail), subst:[x, head, tail] → [y, a0, y1]; ['a] → [tord]

Additional information

Used simplifier rules: ⊦ x ≠ [] → (head = x.head ∧ tail = x.tail ↔ x = head + tail); (elim) Used terms: y, a0, y1

Proof step 17 from proof eq (previous step) (back to tree)

ins≤(a, x0) = ins≤(a, y1), y = a0 + y1, a0 ' + x0 ≠ y, y ≠ [], a0 < a, a0 ∈ a0 ' + ins≤(a, x0), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
ins≤(a, x0) = ins≤(a, y1), x0 ≠ y1, a0 < a, a0 ∈ a0 ' + ins≤(a, y1), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ' + x ≠ []; (s) ⊦ a + x = a ' + x; (s) ⊦ x + y = x + z ↔ y = z; (s)

Proof step 18 from proof eq (previous step) (back to tree)

ins≤(a, x0) = ins≤(a, y1), x0 ≠ y1, a0 < a, a0 ∈ a0 ' + ins≤(a, y1), ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: a, y1

Additional information

Used terms: a, y1

Proof step 19 from proof eq (previous step) (back to tree)

ins≤(a, a0 ' + x0) = ins≤(a, y), a0 ' + x0 ≠ y, ¬ a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
a ' + a0 ' + x0 = ins≤(a, y), a0 ' + x0 ≠ y, a ∈ ins≤(a, y), ¬ a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; a: ⊦ (x + y) + z = x + y + z; from specification list in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list ⊦ a + x = a ' + x; (s)

Proof step 20 from proof eq (previous step) (back to tree)

a ' + a0 ' + x0 = ins≤(a, y), a0 ' + x0 ≠ y, a ∈ ins≤(a, y), ¬ a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ x ≠ [] ∧ x.head < a → ins≤(a, x) = x.head + ins≤(a, x.tail)
Proof step 22:
a ' + a0 ' + x0 = ins≤(a, y), a0 ' + x0 ≠ y, a ∈ ins≤(a, y), ¬ a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦ y ≠ [] ∧ y.head < a

Rule argument

spec: <toplevel>, name:ins-rec, seq: ⊦ x ≠ [] ∧ x.head < a → ins≤(a, x) = x.head + ins≤(a, x.tail), subst:[x, a] → [y, a], <all paths>

Additional information

Used simplifier rules: ⊦ a ' + x = a0 ' + y ↔ a = a0 ∧ x = y; (ws) ⊦ a + x = a ' + x; (s) in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list a: ⊦ (x + y) + z = x + y + z; from specification list nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification weakord[tord] ⊦ ¬ a < a; (s) Used terms: y, a

Proof step 21 from proof eq (previous step) (back to tree)

⊦ x ≠ [] ∧ x.head < a → ins≤(a, x) = x.head + ins≤(a, x.tail)

Lemma ins-rec

Proof step 22 from proof eq (previous step) (back to tree)

a ' + a0 ' + x0 = ins≤(a, y), a0 ' + x0 ≠ y, a ∈ ins≤(a, y), ¬ a0 < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦ y ≠ [] ∧ y.head < a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
a ' + a0 ' + x0 = ins≤(a, y), a0 ' + x0 ≠ y, y ≠ [], a ∈ ins≤(a, y), ¬ a0 < a, ¬ y.head < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Rule argument

Additional information

Used simplifier rules: ins-e: ⊦ ins≤(a, []) = a '; a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ' = a0 ' + x ↔ a = a0 ∧ x = []; (ws)

Proof step 23 from proof eq (previous step) (back to tree)

a ' + a0 ' + x0 = ins≤(a, y), a0 ' + x0 ≠ y, y ≠ [], a ∈ ins≤(a, y), ¬ a0 < a, ¬ y.head < a, ∀ a1, y0. ins≤(a1, x0) ≠ ins≤(a1, y0) ∨ x0 = y0 ⊦

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

spec: list-data, name:elim-add, seq: ⊦ x ≠ [] → (head = x.head ∧ tail = x.tail ↔ x = head + tail), subst:[x, head, tail] → [y, a2, y1]; ['a] → [tord]

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; ⊦ a + x = a ' + x; (s) ⊦ a ' + x = a0 ' + y ↔ a = a0 ∧ x = y; (ws) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ x ≠ [] → (head = x.head ∧ tail = x.tail ↔ x = head + tail); (elim) Used terms: y, a2, y1

Proof Tree for eq (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.