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Proof step 1 from proof ordered-cant-contain-same-element-twice (conclusion) (back to tree)

⊦ ¬ ordered<(x + a ' + y + a ' + z)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof ordered-cant-contain-same-element-twice (previous step) (back to tree)

ordered<(x + a ' + y + a ' + z) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name: ordered-append-split1, seq:⊦ y ≠ [] → (ordered<(x + a ' + y) ↔ ordered<(x + a ') ∧ ordered<(y) ∧ a < y.head), subst:[x, a, y] → [x, a, y + a ' + z], <all paths>

Additional information

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

Proof step 3 from proof ordered-cant-contain-same-element-twice (previous step) (back to tree)

⊦ y ≠ [] → (ordered<(x + a ' + y) ↔ ordered<(x + a ') ∧ ordered<(y) ∧ a < y.head)

Lemma ordered-append-split1

Proof step 4 from proof ordered-cant-contain-same-element-twice (previous step) (back to tree)

y + a ' + z ≠ [], a < (y + a ' + z).head, ordered<(x + a '), ordered<(y + a ' + z) ⊦

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 ⊦ (x + a ' + y).head = (x + a ').head; (s) ⊦ x + a ' + y ≠ []; (s)

Proof step 5 from proof ordered-cant-contain-same-element-twice (previous step) (back to tree)

a < (y + a ').head, ordered<(y + a ' + z), ordered<(x + a ') ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: list-data, name:elim-list-c, seq:x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y, subst:[x, a, y] → [y + a ', a0, y0]

Additional information

Used simplifier rules: ⊦ x + y = [] ↔ x = [] ∧ y = []; (as) ⊦ a ' ≠ []; (s) a: ⊦ (x + y) + z = x + y + z; from specification list x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: y + a ', a0, y0

Proof step 6 from proof ordered-cant-contain-same-element-twice (previous step) (back to tree)

y + a ' = a0 + y0, y + a ' ≠ [], a < a0, ordered<(x + a '), ordered<(y + a ' + z) ⊦

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 ⊦ a ∈ x → a ∈ x + y; (s) ⊦ x.last ∈ x ↔ x ≠ []; (s) ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); (s) in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list inapp-10: ⊦ a ' + x = y + b ' ∧ a ≠ b → a ∈ y ∧ b ∈ x; (forward) from specification list lapp-04: ⊦ a ' + x = y + b ' ∧ a ≠ b → x.last = b; (forward) from specification list-last ⊦ a ∈ b ' ↔ a = b; (s) ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s) ⊦ ¬ a < a; (s) ⊦ a ∈ b ' + x ↔ a = b ∨ a ∈ x; (s) ⊦ x + a ' ≠ []; (s) ⊦ a + x = a ' + x; (s)

Proof step 7 from proof ordered-cant-contain-same-element-twice (previous step) (back to tree)

y + y0.last ' = a0 ' + y0, y0.last ≠ a0, y0 ≠ [], y0.last < a0, ordered<(x + y0.last '), ordered<(y + y0.last ' + z), a0 ∈ y ⊦

User applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: list-last, name:last-elim, seq:x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a ', subst:[x, a, y] → [y0, a, y1]

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a '; (elim) Used terms: y0, a, y1

Proof step 8 from proof ordered-cant-contain-same-element-twice (previous step) (back to tree)

y + a ' = a0 ' + y0, y0 = y1 + a ', a ≠ a0, y0 ≠ [], a < a0, ordered<(y + a ' + z), ordered<(x + a '), a0 ∈ y ⊦

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 ⊦ a ∈ b ' + x ↔ a = b ∨ a ∈ x; (s) in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list ⊦ x + a ' ≠ []; (s) ⊦ # y = # y0 → (x + y = z + y0 ↔ x = z ∧ y = y0); (ws)

Proof step 9 from proof ordered-cant-contain-same-element-twice (previous step) (back to tree)

a ≠ a0, a < a0, ordered<((a0 ' + y1) + a ' + z), ordered<(x + a ') ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name: orderedls-le, seq: ⊦ ¬ a < a0 ∧ a0 ∈ x → ¬ ordered<(a ' + x), subst:[a0, a, x] → [a, a0, y1 + a ' + z], <all paths>

Additional information

Used simplifier rules: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s) ⊦ a ∈ b ' + x ↔ a = b ∨ a ∈ x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem ⊦ ¬ a < a; (s) Used terms: a, a0, y1 + a ' + z

Proof step 10 from proof ordered-cant-contain-same-element-twice (previous step) (back to tree)

⊦ ¬ a < a0 ∧ a0 ∈ x → ¬ ordered<(a ' + x)

Lemma orderedls-le

Proof Tree for ordered-cant-contain-same-element-twice (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