Proof step 1 from proof orderedls-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[m] < x0[n]) ⊦ ordered<(a + x0) ↔ (∀ m, n. m < n ∧ n < #(a + x0) → (a + x0)[m] < (a + x0)[n])

Rule argument

Additional information

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

Proof step 2 from proof orderedls-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[m] < x0[n]) ⊦ ordered<(a ' + x0) ↔ (∀ m, n. m < n ∧ ¬ # x0 < n → (a ' + x0)[m] < (a ' + x0)[n])

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 orderedls-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.head < x.tail.head ∧ ordered<(x.tail))

• Proof step 5:

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

Rule argument

Additional information

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

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

Lemma ols-rec

Proof step 5 from proof orderedls-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[m] < x0[n]) ⊦ a < x0.head ∧ ordered<(x0) ↔ (∀ m, n. m < n ∧ ¬ # x0 < n → (a ' + x0)[m] < (a ' + x0)[n])

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 ' ≠ []; (s) ⊦ x + y = [] ↔ x = [] ∧ y = []; (s)

Proof step 6 from proof orderedls-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[m] < x0[n]) ⊦ a < a0 ∧ ordered<(x0) ↔ (∀ m, n. m < n ∧ ¬ # x0 < n → (a ' + x0)[m] < (a ' + x0)[n])

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

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 ' ≠ []; (s) ⊦ x + y = [] ↔ x = [] ∧ y = []; (s)

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

User applied rule case distinction

And obtained the following new goals

• Proof step 9:

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

• Proof step 31:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) 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) ⊦ m < n → n ≠ 0; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

  ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]

• Proof step 12:

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

• Proof step 22:

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

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a ' + a0 ' + y, m

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

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

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)[0] = a; (s) ⊦ 0 ≤ m; (s) ⊦ 0 < m ↔ m ≠ 0; (s) ⊦ m < n → n ≠ 0; (s) lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic ⊦ a ' ≠ []; (s) ⊦ x + y = [] ↔ x = [] ∧ y = []; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

  ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]

• Proof step 15:

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

Rule argument

Additional information

Used simplifier rules: ⊦ x + y = [] ↔ x = [] ∧ y = []; (s) ⊦ a ' ≠ []; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a ' + a0 ' + y, n

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

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

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).tail = x; (s) ⊦ a ' ≠ []; (s) ⊦ x + y = [] ↔ x = [] ∧ y = []; (s)

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 17:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

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

Rule argument

Additional information

Used simplifier rules: nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification weakord 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 18 from proof orderedls-all (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 19:

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

Rule argument

Additional information

Used simplifier rules: nat freely generated by 0, +1;; ⊦ 0 ≤ m; (s) ⊦ (a ' + x)[0] = a; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ ¬ a < a; (s)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

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

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification weakord f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic a: ⊦ (x + y) + z = x + y + z; from specification list c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ (a ' + x)[n + 1] = x[n]; (s) ⊦ m +1 = m + 1; (s) ⊦ m + 1 ≤ n ↔ m < n; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 21:

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < # x → (x[m] ∈ x ↔ true); (ss) Used terms: y[m1], y, a0

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

Lemma orderedls-ls

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

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

Rule argument

Additional information

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

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 24:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

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

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) ⊦ # x = 0 ↔ x = []; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ m + 1 ≠ 0; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]

• Proof step 27:

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

Rule argument

Additional information

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

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

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

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ (a ' + x).tail = x; (s) ⊦ m < n → n ≠ 0; (s) ⊦ a ' ≠ []; (s) ⊦ x + y = [] ↔ x = [] ∧ y = []; (s)

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 29:

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

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s) n0 < n ⊦ ¬ m + (* n) < (* n0); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat 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 + n < m + n0 ↔ n < n0; (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 30 from proof orderedls-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: m0, m1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

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

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 33:

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

• Proof step 35:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

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

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 34 from proof orderedls-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 + n0 ↔ n < n0; (s) ⊦ m ≤ m0 ∧ n ≤ n0 → ¬ m0 + n0 < m + n; (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 35 from proof orderedls-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

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

Rule argument

Additional information

Proof step 36 from proof orderedls-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