Proof step 1 from proof blastN-len (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  # butlastn(n, x) ≠ # x - n, n ≤ # x ⊦

Rule argument

Additional information

Proof step 2 from proof blastN-len (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  # butlastn(n, x) ≠ # x - n, n ≤ # x ⊦

Rule argument

Additional information

Proof step 3 from proof blastN-len (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 4:

  # x = n0 + n, # butlastn(n, x) ≠ n0, n ≤ # x ⊦

Rule argument

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x, n, n0

Proof step 4 from proof blastN-len (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  # x = n0 + n, # butlastn(n, x) ≠ n0 ⊦

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 ≤ m + n; (s)

Proof step 5 from proof blastN-len (previous step) (back to tree)

User applied rule induction

And obtained the following new goals

• Proof step 6:

  # x = n0 + n, # butlastn(n, x) ≠ n0, ∀ x0, n0, n. # x0 < # x → # x0 = n0 + n ∧ # butlastn(n, x0) ≠ n0 → false ⊦

Rule argument

Additional information

Used simplifier rules: <; (wellfounded) Used terms: # x

Proof step 6 from proof blastN-len (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 7:

  # x = n0 + n, # butlastn(n, x) ≠ n0, x ≠ [], ∀ x0, n0, n. # x0 < # x → # x0 = n0 + n ∧ # butlastn(n, x0) ≠ n0 → false ⊦

Rule argument

Additional information

Used simplifier rules: butlastN-base: ⊦ butlastn(0, x) = x; ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ # [] = 0; (s) Used formulas: x = []

Proof step 7 from proof blastN-len (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 8:

  ⊦ x ≠ [] → (∃ y, a. x = y + a ')

• Proof step 9:

  # x = n0 + n, # butlastn(n, x) ≠ n0, x ≠ [], x ≠ [] → (∃ y, a. x = y + a '), ∀ x0, n0, n. # x0 < # x → # x0 = n0 + n ∧ # butlastn(n, x0) ≠ n0 → false ⊦

Rule argument

Additional information

Used terms: x

Proof step 8 from proof blastN-len (previous step) (back to tree)

A spec-lemma

Proof step 9 from proof blastN-len (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  # y + 1 = n0 + n, # butlastn(n, y + a ') ≠ n0, ∀ x0, n0, n. # x0 < #(y + a ') → # x0 = n0 + n ∧ # butlastn(n, x0) ≠ n0 → false ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ a0 ' ↔ a = a0; (s) ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s) ⊦ x + a ' ≠ []; (s) ⊦ # a ' = 0 +1; (s) ⊦ #(x + y) = # x + # y; (s) in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list

Proof step 10 from proof blastN-len (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

  ⊦ n ≠ 0 → butlastn(n, x + a ') = butlastn(n -1, x)

• Proof step 12:

  # y + 1 = n0 + n, # butlastn(n -1, y) ≠ n0, n ≠ 0, ∀ x0, n0, n. # x0 < #(y + a ') → # x0 = n0 + n ∧ # butlastn(n, x0) ≠ n0 → false ⊦

Rule argument

Additional information

Used simplifier rules: butlastN-base: ⊦ butlastn(0, x) = x; ⊦ 0 + m = m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ #(x + y) = # x + # y; (s) ⊦ # a ' = 0 +1; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a, y, n

Proof step 11 from proof blastN-len (previous step) (back to tree)

Lemma blastN-rec

Proof step 12 from proof blastN-len (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  # y + 1 = n0 + n, # butlastn(n - 1, y) ≠ n0, n ≠ 0, ∀ x0, n0, n. # x0 < #(y + a ') → # x0 = n0 + n ∧ # butlastn(n, x0) ≠ n0 → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n -1 = n - 1; (s)

Proof step 13 from proof blastN-len (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 14:

  # y + 1 = n0 + n, n = n1 + 1, # butlastn(n1, y) ≠ n0, n ≠ 0, ∀ x0, n0, n. # x0 < #(y + a ') → # x0 = n0 + n ∧ # butlastn(n, x0) ≠ n0 → false ⊦

Rule argument

Additional information

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

Proof step 14 from proof blastN-len (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  # y = n0 + n1, # butlastn(n1, y) ≠ n0, ∀ x0, n0, n. # x0 < #(y + a ') → # x0 = n0 + n ∧ # butlastn(n, x0) ≠ n0 → false ⊦

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 + 1 ≠ 0; (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 15 from proof blastN-len (previous step) (back to tree)

User applied rule apply induction

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ #(x + y) = # x + # y; (s) ⊦ # a ' = 0 +1; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ n < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: y, n0, n1