Proof step 1 from proof nodups-butlastn (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(dups(butlastn(n, x)) ↔ false), n ≤ # x, ¬ dups(x) ⊦

Rule argument

Additional information

Proof step 2 from proof nodups-butlastn (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  dups(butlastn(n, x)), n ≤ # x, ¬ dups(x) ⊦

Rule argument

Additional information

Proof step 3 from proof nodups-butlastn (previous step) (back to tree)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

• Proof step 4:

  dups(butlastn(n, a + x0)), n ≤ #(a + x0), ¬ dups(a + x0), ∀ n. ¬(dups(butlastn(n, x0)) ∧ n ≤ # x0 ∧ ¬ dups(x0)) ⊦

Rule argument

Additional information

Used simplifier rules: butlastN-base: ⊦ butlastn(0, x) = x; list('a) freely generated by [], +;; ⊦ # [] = 0; (s) ⊦ m ≤ 0 ↔ m = 0; (s) ⊦ ¬ dups([]); (s)

Proof step 4 from proof nodups-butlastn (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  dups(butlastn(n, a ' + x0)), n ≤ # x0 + 1, ¬ dups(x0), ¬ a ∈ x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ dups(a ' + x) ↔ a ∈ x ∨ dups(x); (s) ⊦ a + x = a ' + x; (s) ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s)

Proof step 5 from proof nodups-butlastn (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

  ⊦ n ≤ # x → butlastn(n, a ' + x) = a ' + butlastn(n, x)

• Proof step 7:

  dups(butlastn(n, a ' + x0)), n ≤ # x0 + 1, ¬ dups(x0), ¬ a ∈ x0, ¬ n ≤ # x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦

• Proof step 13:

  dups(a ' + butlastn(n, x0)), n ≤ # x0 + 1, n ≤ # x0, ¬ dups(x0), ¬ a ∈ x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦

Rule argument

Additional information

Used terms: x0, a, n

Proof step 6 from proof nodups-butlastn (previous step) (back to tree)

Lemma blastN-rec-01

Proof step 7 from proof nodups-butlastn (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  dups(butlastn(# x0, (a ' + x0).butlast)), ¬ dups(x0), ¬ a ∈ x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦

Rule argument

Additional information

Used simplifier rules: butlastN-rec: ⊦ butlastn(n + 1, x) = butlastn(n, x.butlast); c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) ⊦ m + 1 ≠ 0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s) ⊦ m ≤ n ↔ ¬ n < m; (ss)

Proof step 8 from proof nodups-butlastn (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

  ⊦ y ≠ [] → (x + y).butlast = x + y.butlast

• Proof step 10:

  x0 ≠ [], dups(butlastn(# x0, a ' + x0.butlast)), ¬ dups(x0), ¬ a ∈ x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦

Rule argument

Additional information

Used simplifier rules: butlast-one: ⊦ a '.butlast = []; butlastN-base: ⊦ butlastn(0, x) = x; ⊦ # [] = 0; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ ¬ dups([]); (s) Used terms: x0, a '

Proof step 9 from proof nodups-butlastn (previous step) (back to tree)

Lemma bapp

Proof step 10 from proof nodups-butlastn (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  x0 ≠ [], dups(butlastn(# x0, a ' + x0.butlast)), ¬ dups(x0), ¬ a ∈ x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦

Rule argument

Additional information

Proof step 11 from proof nodups-butlastn (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 12:

  n = # x ⊦ butlastn(n, x) = []

Rule argument

Additional information

Used simplifier rules: lb: ⊦ x ≠ [] → # x.butlast = # x - 1; lb: ⊦ x ≠ [] → # x.butlast = # x - 1; ⊦ #(a ' + x) = (# x)+1; (s) ⊦ m +1 = m + 1; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < 1 ↔ m = 0; (s) ⊦ # x = 0 ↔ x = []; (s) n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ¬ dups([]); (s) Used terms: a ' + x0.butlast, # x0

Proof step 12 from proof nodups-butlastn (previous step) (back to tree)

Lemma blastN-all-01

Proof step 13 from proof nodups-butlastn (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  n ≤ # x0, n ≤ # x0 + 1, ¬ dups(x0), ¬ a ∈ x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0, a ∈ butlastn(n, x0) ∨ dups(butlastn(n, x0)) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ dups(a ' + x) ↔ a ∈ x ∨ dups(x); (s)

Proof step 14 from proof nodups-butlastn (previous step) (back to tree)

Heuristic cut applied rule cut formula

And obtained the following new goals

• Proof step 15:

  a ∈ butlastn(n, x0), n ≤ # x0 + 1, n ≤ # x0, ¬ dups(x0), ¬ a ∈ x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0, a ∈ butlastn(n, x0) ∨ dups(butlastn(n, x0)) ⊦

• Proof step 29:

  n ≤ # x0 + 1, n ≤ # x0, ¬ dups(x0), ¬ a ∈ butlastn(n, x0), ¬ a ∈ x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0, a ∈ butlastn(n, x0) ∨ dups(butlastn(n, x0)) ⊦

Rule argument

Additional information

Used formulas: a ∈ butlastn(n, x0)

Proof step 15 from proof nodups-butlastn (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  a ∈ butlastn(n, x0), n ≤ # x0, n ≤ # x0 + 1, ¬ dups(x0), ¬ a ∈ x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦

Rule argument

Additional information

Proof step 16 from proof nodups-butlastn (previous step) (back to tree)

User applied rule weakening

And obtained the following new goals

• Proof step 17:

  a ∈ butlastn(n, x0), n ≤ # x0, ¬ dups(x0), ¬ a ∈ x0 ⊦

Rule argument

Additional information

Proof step 17 from proof nodups-butlastn (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 18:

  a ∈ butlastn(n, a0 + x), n ≤ #(a0 + x), ¬ dups(a0 + x), ¬ a ∈ a0 + x, ∀ a, n. ¬(a ∈ butlastn(n, x) ∧ n ≤ # x ∧ ¬ dups(x) ∧ ¬ a ∈ x) ⊦

Rule argument

Additional information

Used simplifier rules: butlastN-base: ⊦ butlastn(0, x) = x; list('a) freely generated by [], +;; ⊦ # [] = 0; (s) ⊦ m ≤ 0 ↔ m = 0; (s) ⊦ ¬ a ∈ []; (s)

Proof step 18 from proof nodups-butlastn (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  a ≠ a0, a ∈ butlastn(n, a0 ' + x), n ≤ # x + 1, ¬ dups(a0 ' + x), ¬ a ∈ x, ∀ a1, n0. ¬ a1 ∈ butlastn(n0, x) ∨ ¬ n0 ≤ # x ∨ dups(x) ∨ a1 ∈ x ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ a0 ' + x ↔ ¬(a ≠ a0 ∧ ¬ a ∈ x); (ss) ⊦ a + x = a ' + x; (s) ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s)

Proof step 19 from proof nodups-butlastn (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 20:

  ⊦ n ≤ # x → butlastn(n, a ' + x) = a ' + butlastn(n, x)

• Proof step 21:

  a ≠ a0, a ∈ butlastn(n, a0 ' + x), n ≤ # x + 1, ¬ dups(a0 ' + x), ¬ a ∈ x, ¬ n ≤ # x, ∀ a1, n0. ¬ a1 ∈ butlastn(n0, x) ∨ ¬ n0 ≤ # x ∨ dups(x) ∨ a1 ∈ x ⊦

• Proof step 27:

  a ≠ a0, a ∈ a0 ' + butlastn(n, x), n ≤ # x + 1, n ≤ # x, ¬ dups(a0 ' + x), ¬ a ∈ x, ∀ a1, n0. ¬ a1 ∈ butlastn(n0, x) ∨ ¬ n0 ≤ # x ∨ dups(x) ∨ a1 ∈ x ⊦

Rule argument

Additional information

Used terms: x, a0, n

Proof step 20 from proof nodups-butlastn (previous step) (back to tree)

Lemma blastN-rec-01

Proof step 21 from proof nodups-butlastn (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  a ≠ a0, a ∈ butlastn(# x, (a0 ' + x).butlast), ¬ dups(a0 ' + x), ¬ a ∈ x, ∀ a1, n0. ¬ a1 ∈ butlastn(n0, x) ∨ ¬ n0 ≤ # x ∨ dups(x) ∨ a1 ∈ x ⊦

Rule argument

Additional information

Used simplifier rules: butlastN-rec: ⊦ butlastn(n + 1, x) = butlastn(n, x.butlast); c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) ⊦ m + 1 ≠ 0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s) ⊦ m ≤ n ↔ ¬ n < m; (ss)

Proof step 22 from proof nodups-butlastn (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 23:

  ⊦ y ≠ [] → (x + y).butlast = x + y.butlast

• Proof step 24:

  a ≠ a0, x ≠ [], a ∈ butlastn(# x, a0 ' + x.butlast), ¬ dups(a0 ' + x), ¬ a ∈ x, ∀ a1, n0. ¬ a1 ∈ butlastn(n0, x) ∨ ¬ n0 ≤ # x ∨ dups(x) ∨ a1 ∈ x ⊦

Rule argument

Additional information

Used simplifier rules: butlast-one: ⊦ a '.butlast = []; butlastN-base: ⊦ butlastn(0, x) = x; ⊦ # [] = 0; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ ¬ a ∈ []; (s) Used terms: x, a0 '

Proof step 23 from proof nodups-butlastn (previous step) (back to tree)

Lemma bapp

Proof step 24 from proof nodups-butlastn (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  a ≠ a0, x ≠ [], a ∈ butlastn(# x, a0 ' + x.butlast), ¬ dups(a0 ' + x), ¬ a ∈ x, ∀ a1, n0. ¬ a1 ∈ butlastn(n0, x) ∨ ¬ n0 ≤ # x ∨ dups(x) ∨ a1 ∈ x ⊦

Rule argument

Additional information

Proof step 25 from proof nodups-butlastn (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  n = # x ⊦ butlastn(n, x) = []

Rule argument

Additional information

Used simplifier rules: lb: ⊦ x ≠ [] → # x.butlast = # x - 1; lb: ⊦ x ≠ [] → # x.butlast = # x - 1; ⊦ #(a ' + x) = (# x)+1; (s) ⊦ m +1 = m + 1; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < 1 ↔ m = 0; (s) ⊦ # x = 0 ↔ x = []; (s) n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ¬ a ∈ []; (s) Used terms: a0 ' + x.butlast, # x

Proof step 26 from proof nodups-butlastn (previous step) (back to tree)

Lemma blastN-all-01

Proof step 27 from proof nodups-butlastn (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  a ≠ a0, a ∈ butlastn(n, x), n ≤ # x, n ≤ # x + 1, ¬ dups(x), ¬ a ∈ x, ¬ a0 ∈ x, ∀ a1, n0. ¬ a1 ∈ butlastn(n0, x) ∨ ¬ n0 ≤ # x ∨ a1 ∈ x ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ dups(a ' + x) ↔ a ∈ x ∨ dups(x); (s) ⊦ a ≠ a0 → (a ∈ a0 ' + x ↔ a ∈ x); (s)

Proof step 28 from proof nodups-butlastn (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: a, n

Proof step 29 from proof nodups-butlastn (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  dups(butlastn(n, x0)), n ≤ # x0, n ≤ # x0 + 1, ¬ dups(x0), ¬ a ∈ butlastn(n, x0), ¬ a ∈ x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦

Rule argument

Additional information

Proof step 30 from proof nodups-butlastn (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n