Proof step 1 from proof max-append (conclusion) (back to tree)

⊦ x + x0 ≠ [] → max(x + x0) = max(x) ∨ max(x + x0) = max(x0)

System applied rule normalize

And obtained the following new goals

Proof step 2:
x + x0 ≠ [], max(x + x0) ≠ max(x), max(x + x0) ≠ max(x0) ⊦

Rule argument

Additional information

Proof step 2 from proof max-append (previous step) (back to tree)

x + x0 ≠ [], max(x + x0) ≠ max(x), max(x + x0) ≠ max(x0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x + x0 ≠ [], max(x + x0) ≠ max(x), max(x + x0) ≠ max(x0) ⊦

Rule argument

Additional information

Proof step 3 from proof max-append (previous step) (back to tree)

x + x0 ≠ [], max(x + x0) ≠ max(x), max(x + x0) ≠ max(x0) ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 4:
(a + x1) + x0 ≠ [], max((a + x1) + x0) ≠ max(a + x1), max((a + x1) + x0) ≠ max(x0), ∀ x0. ¬(x1 + x0 ≠ [] ∧ max(x1 + x0) ≠ max(x1) ∧ max(x1 + x0) ≠ max(x0)) ⊦

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;; ⊦ [] + x = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 4 from proof max-append (previous step) (back to tree)

(a + x1) + x0 ≠ [], max((a + x1) + x0) ≠ max(a + x1), max((a + x1) + x0) ≠ max(x0), ∀ x0. ¬(x1 + x0 ≠ [] ∧ max(x1 + x0) ≠ max(x1) ∧ max(x1 + x0) ≠ max(x0)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
max((a ' + x1) + x0) ≠ max(a ' + x1), max((a ' + x1) + x0) ≠ max(x0), ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

Additional information

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

Proof step 5 from proof max-append (previous step) (back to tree)

max((a ' + x1) + x0) ≠ max(a ' + x1), max((a ' + x1) + x0) ≠ max(x0), ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 7:
x1 + x0 ≠ [], (max(x1 + x0) < a ⊃ a;max(x1 + x0)) ≠ max(a ' + x1), (max(x1 + x0) < a ⊃ a;max(x1 + x0)) ≠ max(x0), ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

spec: <toplevel>, name:max-rec, seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)), subst:[x, a] → [x1 + x0, a], <all paths>

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a; ⊦ x + y = [] ↔ x = [] ∧ y = []; (as) ⊦ [] + x = x; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x1 + x0, a

Proof step 6 from proof max-append (previous step) (back to tree)

⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))

Proof step 7 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], (max(x1 + x0) < a ⊃ a;max(x1 + x0)) ≠ max(a ' + x1), (max(x1 + x0) < a ⊃ a;max(x1 + x0)) ≠ max(x0), ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
x1 + x0 ≠ [], (max(x1 + x0) < a ⊃ a;max(x1 + x0)) ≠ max(a ' + x1), (max(x1 + x0) < a ⊃ a;max(x1 + x0)) ≠ max(x0), ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

Additional information

Proof step 8 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], (max(x1 + x0) < a ⊃ a;max(x1 + x0)) ≠ max(a ' + x1), (max(x1 + x0) < a ⊃ a;max(x1 + x0)) ≠ max(x0), ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 9:
x1 + x0 ≠ [], (max(x1 + x0) < a ⊃ a;max(x1 + x0)) ≠ max(x0), a ≠ max(a ' + x1), max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦
Proof step 14:
x1 + x0 ≠ [], (max(x1 + x0) < a ⊃ a;max(x1 + x0)) ≠ max(x0), max(x1 + x0) ≠ max(a ' + x1), ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

pos: left 2

Additional information

Proof step 9 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], (max(x1 + x0) < a ⊃ a;max(x1 + x0)) ≠ max(x0), a ≠ max(a ' + x1), max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
x1 + x0 ≠ [], a ≠ max(a ' + x1), a ≠ max(x0), max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

Additional information

Proof step 10 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], a ≠ max(a ' + x1), a ≠ max(x0), max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 12:
x1 + x0 ≠ [], a ≠ (max(x1) < a ⊃ a;max(x1)), a ≠ max(x0), x1 ≠ [], max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

spec: <toplevel>, name:max-rec, seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)), subst:[x, a] → [x1, a], <all paths>

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a; ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x1, a

Proof step 11 from proof max-append (previous step) (back to tree)

⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))

Proof step 12 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], a ≠ (max(x1) < a ⊃ a;max(x1)), a ≠ max(x0), x1 ≠ [], max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
x1 + x0 ≠ [], a ≠ (max(x1) < a ⊃ a;max(x1)), a ≠ max(x0), x1 ≠ [], max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

Additional information

Proof step 13 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], a ≠ (max(x1) < a ⊃ a;max(x1)), a ≠ max(x0), x1 ≠ [], max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 2

Additional information

Used simplifier rules: max-ge: ⊦ x ≠ [] → (max(x + x0) < max(x) ↔ false); ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss) transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 14 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], (max(x1 + x0) < a ⊃ a;max(x1 + x0)) ≠ max(x0), max(x1 + x0) ≠ max(a ' + x1), ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
x1 + x0 ≠ [], max(x1 + x0) ≠ max(a ' + x1), max(x1 + x0) ≠ max(x0), ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

Additional information

Proof step 15 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], max(x1 + x0) ≠ max(a ' + x1), max(x1 + x0) ≠ max(x0), ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 17:
x1 + x0 ≠ [], max(x1 + x0) ≠ (max(x1) < a ⊃ a;max(x1)), max(x1 + x0) ≠ max(x0), x1 ≠ [], ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

spec: <toplevel>, name:max-rec, seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)), subst:[x, a] → [x1, a], <all paths>

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a; ⊦ x + [] = x; (s) ⊦ [] + x = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x1, a

Proof step 16 from proof max-append (previous step) (back to tree)

⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))

Lemma max-rec

Proof step 17 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], max(x1 + x0) ≠ (max(x1) < a ⊃ a;max(x1)), max(x1 + x0) ≠ max(x0), x1 ≠ [], ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
x1 + x0 ≠ [], max(x1 + x0) ≠ (max(x1) < a ⊃ a;max(x1)), max(x1 + x0) ≠ max(x0), x1 ≠ [], ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

Additional information

Proof step 18 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], max(x1 + x0) ≠ (max(x1) < a ⊃ a;max(x1)), max(x1 + x0) ≠ max(x0), x1 ≠ [], ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 19:
x1 + x0 ≠ [], max(x1 + x0) ≠ a, max(x1 + x0) ≠ max(x0), x1 ≠ [], max(x1) < a, ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦
Proof step 21:
x1 + x0 ≠ [], max(x1 + x0) ≠ max(x1), max(x1 + x0) ≠ max(x0), x1 ≠ [], ¬ max(x1) < a, ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

pos: left 2

Additional information

Proof step 19 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], max(x1 + x0) ≠ a, max(x1 + x0) ≠ max(x0), x1 ≠ [], max(x1) < a, ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
x1 + x0 ≠ [], max(x1 + x0) ≠ a, max(x1 + x0) ≠ max(x0), x1 ≠ [], max(x1) < a, a < max(x1 + x0), max(x1) < max(x1 + x0), ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

Additional information

Used simplifier rules: transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss)

Proof step 20 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], max(x1 + x0) ≠ a, max(x1 + x0) ≠ max(x0), x1 ≠ [], max(x1) < a, a < max(x1 + x0), max(x1) < max(x1 + x0), ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: x0

Additional information

Used simplifier rules: max-ge: ⊦ x ≠ [] → (max(x + x0) < max(x) ↔ false); transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0

Proof step 21 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], max(x1 + x0) ≠ max(x1), max(x1 + x0) ≠ max(x0), x1 ≠ [], ¬ max(x1) < a, ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
x1 + x0 ≠ [], max(x1 + x0) ≠ max(x1), max(x1 + x0) ≠ max(x0), x1 ≠ [], ¬ max(x1) < a, ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Rule argument

Additional information

Proof step 22 from proof max-append (previous step) (back to tree)

x1 + x0 ≠ [], max(x1 + x0) ≠ max(x1), max(x1 + x0) ≠ max(x0), x1 ≠ [], ¬ max(x1) < a, ¬ max(x1 + x0) < a, ∀ x. x1 + x = [] ∨ max(x1 + x) = max(x1) ∨ max(x1 + x) = max(x) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: x0

Additional information

Used terms: x0

Proof Tree for max-append (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.