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Proof step 1 from proof ordered-toList-ordered (conclusion) (back to tree)

⊦ ordered≤(t) → ordered≤(t.toList)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof ordered-toList-ordered (previous step) (back to tree)

ordered≤(t), ¬ ordered≤(t.toList) ⊦

User applied rule structural induction

And obtained the following new goals

Rule argument

term: t

Additional information

Used simplifier rules: ordered-base: ⊦ ordered≤(leaf); toList-base: ⊦ leaf.toList = []; tree freely generated by leaf, branch;; ⊦ ordered≤([]); (s)

Proof step 3 from proof ordered-toList-ordered (previous step) (back to tree)

ordered≤(branch(t0, a, t1)), ¬ ordered≤(branch(t0, a, t1).toList) ⊦ ordered≤(t0) ∧ ¬ ordered≤(t0.toList), ordered≤(t1) ∧ ¬ ordered≤(t1.toList)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ordered-rec: ⊦ ordered≤(branch(t0, a, t1)) ↔ ordered≤(t0) ∧ t0 ≤ a ∧ a ≤ t1 ∧ ordered≤(t1); toList-rec: ⊦ branch(t0, a, t1).toList = t0.toList + (a + t1.toList); ⊦ a + x = a ' + x; (s)

Proof step 4 from proof ordered-toList-ordered (previous step) (back to tree)

ordered≤(t0), ordered≤(t1), ordered≤(t1.toList), ordered≤(t0.toList), t0 ≤ a, a ≤ t1, ¬ ordered≤(t0.toList + a ' + t1.toList) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: olist, name:ordered-append, seq:⊦ ordered≤(x) ∧ ordered≤(x0) ∧ (x ≠ [] ∧ x0 ≠ [] → ¬ x0.head < x.last) → ordered≤(x + x0), subst:[x, x0] → [t0.toList, a ' + t1.toList], <all paths>

Additional information

Used terms: t0.toList, a ' + t1.toList

Proof step 5 from proof ordered-toList-ordered (previous step) (back to tree)

⊦ ordered≤(x) ∧ ordered≤(x0) ∧ (x ≠ [] ∧ x0 ≠ [] → ¬ x0.head < x.last) → ordered≤(x + x0)

A spec-lemma

Proof step 6 from proof ordered-toList-ordered (previous step) (back to tree)

ordered≤(t0.toList), ordered≤(t1.toList), ordered≤(t1), ordered≤(t0), a ≤ t1, t0 ≤ a, ¬ ordered≤(t0.toList + a ' + t1.toList) ⊦ ordered≤(t0.toList) ∧ ordered≤(a ' + t1.toList) ∧ (t0.toList ≠ [] ∧ a ' + t1.toList ≠ [] → ¬ (a ' + t1.toList).head < t0.toList.last)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: toList-base: ⊦ leaf.toList = []; tree-le-elem-toList-last: t ≠ leaf ⊦ t ≤ a → ¬ a < t.toList.last; a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ (a ' + x).head = a; (s) ⊦ a ' + x ≠ []; (s)

Proof step 7 from proof ordered-toList-ordered (previous step) (back to tree)

ordered≤(t0), ordered≤(t1), ordered≤(t1.toList), ordered≤(t0.toList), t0 ≤ a, a ≤ t1, ¬ ordered≤(t0.toList + a ' + t1.toList), ¬ ordered≤(a ' + t1.toList) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: olist, name:ordered-append, seq:⊦ ordered≤(x) ∧ ordered≤(x0) ∧ (x ≠ [] ∧ x0 ≠ [] → ¬ x0.head < x.last) → ordered≤(x + x0), subst:[x, x0] → [a ', t1.toList], <all paths>

Additional information

Used terms: a ', t1.toList

Proof step 8 from proof ordered-toList-ordered (previous step) (back to tree)

⊦ ordered≤(x) ∧ ordered≤(x0) ∧ (x ≠ [] ∧ x0 ≠ [] → ¬ x0.head < x.last) → ordered≤(x + x0)

A spec-lemma

Proof step 9 from proof ordered-toList-ordered (previous step) (back to tree)

ordered≤(t0.toList), ordered≤(t1.toList), ordered≤(t1), ordered≤(t0), a ≤ t1, t0 ≤ a, ¬ ordered≤(t0.toList + a ' + t1.toList), ¬ ordered≤(a ' + t1.toList) ⊦ ordered≤(a ') ∧ ordered≤(t1.toList) ∧ (a ' ≠ [] ∧ t1.toList ≠ [] → ¬ t1.toList.head < a '.last)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ a '.last = a; (s) ⊦ a ' ≠ []; (s) ⊦ ordered≤(a '); (s)

Proof step 10 from proof ordered-toList-ordered (previous step) (back to tree)

t1.toList ≠ [], t1.toList.head < a, ordered≤(t0), ordered≤(t1), ordered≤(t1.toList), ordered≤(t0.toList), t0 ≤ a, a ≤ t1, ¬ ordered≤(t0.toList + a ' + t1.toList), ¬ ordered≤(a ' + t1.toList) ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: list-data, name:elim-add, seq: ⊦ x ≠ [] → (head = x.head ∧ tail = x.tail ↔ x = head + tail), subst:[x, head, tail] → [t1.toList, a0, x0]

Additional information

Used simplifier rules: ⊦ x ≠ [] → (head = x.head ∧ tail = x.tail ↔ x = head + tail); (elim) Used terms: t1.toList, a0, tail

Proof step 11 from proof ordered-toList-ordered (previous step) (back to tree)

t1.toList = a0 + x0, t1.toList ≠ [], a0 < a, ordered≤(t0.toList), ordered≤(t1.toList), ordered≤(t1), ordered≤(t0), a ≤ t1, t0 ≤ a, ¬ ordered≤(t0.toList + a ' + t1.toList), ¬ ordered≤(a ' + t1.toList) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 12 from proof ordered-toList-ordered (previous step) (back to tree)

t1.toList = a0 ' + x0, t1.toList ≠ [], a0 < a, ordered≤(t0.toList), ordered≤(t1.toList), ordered≤(t1), ordered≤(t0), a0 ∈ t1.toList, a ≤ t1, t0 ≤ a, ¬ ordered≤(t0.toList + a ' + t1.toList) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:elem-le-tree-in-tree, seq: ⊦ b ∈ t ∧ a ≤ t → ¬ b < a, subst:[t, b, a] → [t1, a0, a], <all paths>

Additional information

Used terms: t1, a0, a

Proof step 13 from proof ordered-toList-ordered (previous step) (back to tree)

⊦ b ∈ t ∧ a ≤ t → ¬ b < a

Lemma elem-le-tree-in-tree

Proof step 14 from proof ordered-toList-ordered (previous step) (back to tree)

t1.toList = a0 ' + x0, t1.toList ≠ [], a0 < a, ordered≤(t0), ordered≤(t1), ordered≤(t1.toList), ordered≤(t0.toList), a0 ∈ t1.toList, t0 ≤ a, a ≤ t1, ¬ ordered≤(t0.toList + a ' + t1.toList) ⊦ a0 ∈ t1 ∧ a ≤ t1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 15 from proof ordered-toList-ordered (previous step) (back to tree)

t1.toList = a0 ' + x0, t1.toList ≠ [], a0 < a, ordered≤(t0.toList), ordered≤(t1.toList), ordered≤(t1), ordered≤(t0), a0 ∈ t1.toList, a ≤ t1, t0 ≤ a, ¬ ordered≤(t0.toList + a ' + t1.toList), ¬ a0 ∈ t1 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:in-tree-in-toList, seq: ⊦ a ∈ t ↔ a ∈ t.toList, subst:[a, t] → [a0, t1], <all paths>

Additional information

Used terms: a0, t1

Proof step 16 from proof ordered-toList-ordered (previous step) (back to tree)

⊦ a ∈ t ↔ a ∈ t.toList

Lemma in-tree-in-toList

Proof Tree for ordered-toList-ordered (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.

the proof tree