Proof step 1 from proof orderedle-all (conclusion) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 2:

  ordered≤(x0) ↔ (∀ m, n. m ≤ n ∧ n < # x0 → ¬ x0[n] < x0[m]) ⊦ ordered≤(a + x0) ↔ (∀ m, n. m ≤ n ∧ n < #(a + x0) → ¬ (a + x0)[n] < (a + x0)[m])

Rule argument

Additional information

Used simplifier rules: le-e: ⊦ ordered≤([]); list('a) freely generated by [], +;; ⊦ # [] = 0; (s) ⊦ ¬ n < 0; (s)

Proof step 2 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ordered≤(x0) ↔ (∀ m, n. m ≤ n ∧ n < # x0 → ¬ x0[n] < x0[m]) ⊦ ordered≤(a ' + x0) ↔ (∀ m, n. m ≤ n ∧ ¬ # x0 < n → ¬ (a ' + x0)[n] < (a ' + x0)[m])

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s)

Proof step 3 from proof orderedle-all (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ x ≠ [] ∧ x.tail ≠ [] → (ordered≤(x) ↔ ¬ x.tail.head < x.head ∧ ordered≤(x.tail))

• Proof step 5:

  a ' + x0 ≠ [], (a ' + x0).tail ≠ [], ordered≤(x0) ↔ (∀ m, n. m ≤ n ∧ n < # x0 → ¬ x0[n] < x0[m]) ⊦ ¬ (a ' + x0).tail.head < (a ' + x0).head ∧ ordered≤((a ' + x0).tail) ↔ (∀ m, n. m ≤ n ∧ ¬ # x0 < n → ¬ (a ' + x0)[n] < (a ' + x0)[m])

Rule argument

Additional information

Used simplifier rules: le-e: ⊦ ordered≤([]); le-o: ⊦ ordered≤(a '); ⊦ a ' + x ≠ []; (s) ⊦ (a ' + x).tail = x; (s) ⊦ ¬ n < 0; (s) ⊦ # [] = 0; (s) ⊦ 0 < m ↔ m ≠ 0; (s) ⊦ m ≤ 0 ↔ m = 0; (s) ⊦ a '[0] = a; (s) ⊦ ¬ a < a; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a ' + x0

Proof step 4 from proof orderedle-all (previous step) (back to tree)

Lemma ole-rec

Proof step 5 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  x0 ≠ [], ordered≤(x0) ↔ (∀ m, n. m ≤ n ∧ n < # x0 → ¬ x0[n] < x0[m]) ⊦ ¬ x0.head < a ∧ ordered≤(x0) ↔ (∀ m, n. m ≤ n ∧ ¬ # x0 < n → ¬ (a ' + x0)[n] < (a ' + x0)[m])

Rule argument

Additional information

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

Proof step 6 from proof orderedle-all (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 7:

  x0 = a0 + y, x0 ≠ [], ordered≤(x0) ↔ (∀ m, n. m ≤ n ∧ n < # x0 → ¬ x0[n] < x0[m]) ⊦ ¬ a0 < a ∧ ordered≤(x0) ↔ (∀ m, n. m ≤ n ∧ ¬ # x0 < n → ¬ (a ' + x0)[n] < (a ' + x0)[m])

Rule argument

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x0, a0, y

Proof step 7 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  ordered≤(a0 ' + y) ↔ (∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m]) ⊦ ¬ a0 < a ∧ ordered≤(a0 ' + y) ↔ (∀ m, n. m ≤ n ∧ ¬ # y + 1 < n → ¬ (a ' + a0 ' + y)[n] < (a ' + a0 ' + y)[m])

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a + x = a ' + x; (s) ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ a ' + x ≠ []; (s)

Proof step 8 from proof orderedle-all (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 9:

  ordered≤(a0 ' + y), ¬ a0 < a, ordered≤(a0 ' + y) ↔ (∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m]) ⊦ ∀ m, n. m ≤ n ∧ ¬ # y + 1 < n → ¬ (a ' + a0 ' + y)[n] < (a ' + a0 ' + y)[m]

• Proof step 29:

  ordered≤(a0 ' + y) ↔ (∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m]), ∀ m, n. m ≤ n ∧ ¬ # y + 1 < n → ¬ (a ' + a0 ' + y)[n] < (a ' + a0 ' + y)[m] ⊦ ¬ a0 < a ∧ ordered≤(a0 ' + y)

Rule argument

Additional information

Proof step 9 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  (a ' + a0 ' + y)[n] < (a ' + a0 ' + y)[m], ordered≤(a0 ' + y), m ≤ # y + 1, n ≤ # y + 1, m ≤ n, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x); app-01: ⊦ ordered≤(x + y) → ordered≤(y); let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 10 from proof orderedle-all (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

  n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]

• Proof step 12:

  n ≠ 0, (a0 ' + y)[n -1] < (a ' + a0 ' + y)[m], ordered≤(a0 ' + y), m ≤ n, n ≤ # y + 1, m ≤ # y + 1, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s) ⊦ m ≤ 0 ↔ m = 0; (s) ⊦ (a ' + x)[0] = a; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ ¬ a < a; (s) Used terms: n, a, a0 ' + y

Proof step 11 from proof orderedle-all (previous step) (back to tree)

A spec-lemma

Proof step 12 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  n ≠ 0, (a0 ' + y)[n - 1] < (a ' + a0 ' + y)[m], ordered≤(a0 ' + y), m ≤ # y + 1, n ≤ # y + 1, m ≤ n, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n -1 = n - 1; (s)

Proof step 13 from proof orderedle-all (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 14:

  n = m0 + 1, n ≠ 0, (a0 ' + y)[m0] < (a ' + a0 ' + y)[m], ordered≤(a0 ' + y), m ≤ n, n ≤ # y + 1, m ≤ # y + 1, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0

Proof step 14 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  (a0 ' + y)[m0] < (a ' + a0 ' + y)[m], ordered≤(a0 ' + y), m ≤ m0 + 1, m ≤ # y + 1, m0 ≤ # y, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 15 from proof orderedle-all (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

  n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]

• Proof step 17:

  m = 0, (a0 ' + y)[m0] < (a ' + a0 ' + y)[m], ordered≤(a0 ' + y), m0 ≤ # y, m ≤ # y + 1, m ≤ m0 + 1, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

• Proof step 25:

  m ≠ 0, (a0 ' + y)[m0] < (a0 ' + y)[m -1], ordered≤(a0 ' + y), m0 ≤ # y, m ≤ # y + 1, m ≤ m0 + 1, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used terms: m, a, a0 ' + y

Proof step 16 from proof orderedle-all (previous step) (back to tree)

A spec-lemma

Proof step 17 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  (a0 ' + y)[m0] < a, ordered≤(a0 ' + y), m0 ≤ # y, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ (a ' + x)[0] = a; (s) ⊦ 0 ≤ m; (s)

Proof step 18 from proof orderedle-all (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

  n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]

• Proof step 20:

  m0 ≠ 0, y[m0 -1] < a, ordered≤(a0 ' + y), m0 ≤ # y, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s) ⊦ (a ' + x)[0] = a; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: m0, a0, y

Proof step 19 from proof orderedle-all (previous step) (back to tree)

A spec-lemma

Proof step 20 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  m0 ≠ 0, y[m0 - 1] < a, ordered≤(a0 ' + y), m0 ≤ # y, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n -1 = n - 1; (s)

Proof step 21 from proof orderedle-all (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 22:

  m0 = m1 + 1, m0 ≠ 0, y[m1] < a, ordered≤(a0 ' + y), m0 ≤ # y, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m0, m1

Proof step 22 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  y ≠ [], m1 < # y, y[m1] < a, ordered≤(a0 ' + y), ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 23 from proof orderedle-all (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 24:

  ⊦ b < a ∧ b ∈ x → ¬ ordered≤(a ' + x)

Rule argument

Additional information

Used simplifier rules: ⊦ m < # x → (x[m] ∈ x ↔ true); (ss) nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification weakord[tord] leq-trans: ⊦ ¬ b < a ∧ ¬ c < b → ¬ c < a; (forward) from specification weakord[tord] Used terms: y[m1], a0, y

Proof step 24 from proof orderedle-all (previous step) (back to tree)

Lemma orderedle-ls

Proof step 25 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  m ≠ 0, (a0 ' + y)[m0] < (a0 ' + y)[m - 1], ordered≤(a0 ' + y), m ≤ m0 + 1, m ≤ # y + 1, m0 ≤ # y, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n -1 = n - 1; (s)

Proof step 26 from proof orderedle-all (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 27:

  m = m1 + 1, m ≠ 0, (a0 ' + y)[m0] < (a0 ' + y)[m1], ordered≤(a0 ' + y), m0 ≤ # y, m ≤ # y + 1, m ≤ m0 + 1, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m1

Proof step 27 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  (a0 ' + y)[m0] < (a0 ' + y)[m1], ordered≤(a0 ' + y), m0 ≤ # y, m1 ≤ m0, m1 ≤ # y, ¬ a0 < a, ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 28 from proof orderedle-all (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: m1, m0

Proof step 29 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  ordered≤(a0 ' + y) ↔ (∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m]), ∀ m, n. m ≤ n ∧ ¬ # y + 1 < n → ¬ (a ' + a0 ' + y)[n] < (a ' + a0 ' + y)[m] ⊦ ¬ a0 < a ∧ ordered≤(a0 ' + y)

Rule argument

Additional information

Proof step 30 from proof orderedle-all (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 31:

  ¬ ordered≤(a0 ' + y), ∀ m, n. m ≤ n ∧ ¬ # y + 1 < n → ¬ (a ' + a0 ' + y)[n] < (a ' + a0 ' + y)[m] ⊦ ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m], ¬ a0 < a ∧ ordered≤(a0 ' + y)

• Proof step 33:

  ordered≤(a0 ' + y), ∀ m, n. m ≤ n ∧ ¬ # y + 1 < n → ¬ (a ' + a0 ' + y)[n] < (a ' + a0 ' + y)[m], ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦ ¬ a0 < a ∧ ordered≤(a0 ' + y)

Rule argument

Additional information

Proof step 31 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

  (a0 ' + y)[n] < (a0 ' + y)[m], m ≤ n, n ≤ # y, m ≤ # y, ¬ ordered≤(a0 ' + y), ∀ m, n. m ≤ n ∧ ¬ # y + 1 < n → ¬ (a ' + a0 ' + y)[n] < (a ' + a0 ' + y)[m] ⊦

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 32 from proof orderedle-all (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ n ≤ n0 ∧ n1 ≤ m → n + n1 ≤ n0 + m; (s) ⊦ m ≤ m0 ∧ n ≤ n0 → ¬ m0 + n0 < m + n; (s) ⊦ m +1 = m + 1; (s) ⊦ (a ' + x)[n + 1] = x[n]; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: m + 1, n +1

Proof step 33 from proof orderedle-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  a0 < a, ordered≤(a0 ' + y), ∀ m, n. m ≤ n ∧ ¬ # y + 1 < n → ¬ (a ' + a0 ' + y)[n] < (a ' + a0 ' + y)[m], ∀ m, n. m ≤ n ∧ ¬ # y < n → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦

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 34 from proof orderedle-all (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n → ¬ m + m0 < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]; (s) ⊦ (a ' + x)[0] = a; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: 0, 1