Proof step 1 from proof at-last (conclusion) (back to tree)

⊦ x ≠ [] → x[# x - 1] = x.last

System applied rule normalize

And obtained the following new goals

Proof step 2:
x[# x - 1] ≠ x.last, x ≠ [] ⊦

Rule argument

Additional information

Proof step 2 from proof at-last (previous step) (back to tree)

x[# x - 1] ≠ x.last, x ≠ [] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x[# x - 1] ≠ x.last, x ≠ [] ⊦

Rule argument

Additional information

Proof step 3 from proof at-last (previous step) (back to tree)

x[# x - 1] ≠ x.last, x ≠ [] ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Proof step 4:
# x = m + 1, # x ≠ 0, x[m] ≠ x.last, x ≠ [] ⊦

Rule argument

spec: nat, name:elim-pred-c, seq:n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1, subst:[n, m] → [# x, m]

Additional information

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

Proof step 4 from proof at-last (previous step) (back to tree)

# x = m + 1, # x ≠ 0, x[m] ≠ x.last, x ≠ [] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# x = m + 1, x[m] ≠ x.last, x ≠ [] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)

Proof step 5 from proof at-last (previous step) (back to tree)

# x = m + 1, x[m] ≠ x.last, x ≠ [] ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 6:
#(a + x0) = m + 1, a + x0 ≠ [], (a + x0)[m] ≠ (a + x0).last, ∀ m. ¬(# x0 = m + 1 ∧ x0[m] ≠ x0.last ∧ x0 ≠ []) ⊦

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;;

Proof step 6 from proof at-last (previous step) (back to tree)

#(a + x0) = m + 1, a + x0 ≠ [], (a + x0)[m] ≠ (a + x0).last, ∀ m. ¬(# x0 = m + 1 ∧ x0[m] ≠ x0.last ∧ x0 ≠ []) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
(a ' + x0)[# x0] ≠ (a ' + x0).last, ∀ m0. # x0 ≠ m0 + 1 ∨ x0[m0] = x0.last ∨ x0 = [] ⊦

Rule argument

Additional information

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

Proof step 7 from proof at-last (previous step) (back to tree)

(a ' + x0)[# x0] ≠ (a ' + x0).last, ∀ m0. # x0 ≠ m0 + 1 ∨ x0[m0] = x0.last ∨ x0 = [] ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 8:
(a ' + x0)[# x0] ≠ (a ' + x0).last, x0 ≠ [], ∀ m0. # x0 ≠ m0 + 1 ∨ x0[m0] = x0.last ∨ x0 = [] ⊦

Rule argument

fma: x0 = []

Additional information

Used simplifier rules: one-03: ⊦ a '[0] = a; ⊦ # [] = 0; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a '.last = a; (s) Used formulas: x0 = []

Proof step 8 from proof at-last (previous step) (back to tree)

(a ' + x0)[# x0] ≠ (a ' + x0).last, x0 ≠ [], ∀ m0. # x0 ≠ m0 + 1 ∨ x0[m0] = x0.last ∨ x0 = [] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
(a ' + x0)[# x0] ≠ x0.last, x0 ≠ [], ∀ m0. # x0 ≠ m0 + 1 ∨ x0[m0] = x0.last ⊦

Rule argument

Additional information

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

Proof step 9 from proof at-last (previous step) (back to tree)

(a ' + x0)[# x0] ≠ x0.last, x0 ≠ [], ∀ m0. # x0 ≠ m0 + 1 ∨ x0[m0] = x0.last ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]
Proof step 11:
# x0 ≠ 0, a ' + x0 ≠ [], (a ' + x0).tail[# x0 - 1] ≠ x0.last, x0 ≠ [], ∀ m0. # x0 ≠ m0 + 1 ∨ x0[m0] = x0.last ⊦

Rule argument

spec: <toplevel>, name:get-rec, seq: ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1], subst:[n, x] → [# x0, a ' + x0], <all paths>

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) ⊦ x + y = [] ↔ x = [] ∧ y = []; (as) ⊦ a ' ≠ []; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: # x0, a ' + x0

Proof step 10 from proof at-last (previous step) (back to tree)

⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]

Lemma get-rec

Proof step 11 from proof at-last (previous step) (back to tree)

# x0 ≠ 0, a ' + x0 ≠ [], (a ' + x0).tail[# x0 - 1] ≠ x0.last, x0 ≠ [], ∀ m0. # x0 ≠ m0 + 1 ∨ x0[m0] = x0.last ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
x0[# x0 - 1] ≠ x0.last, x0 ≠ [], ∀ m0. # x0 ≠ m0 + 1 ∨ x0[m0] = x0.last ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ (a ' + x).tail = x; (s) ⊦ a ' + x ≠ []; (s) ⊦ # x = 0 ↔ x = []; (s)

Proof step 12 from proof at-last (previous step) (back to tree)

x0[# x0 - 1] ≠ x0.last, x0 ≠ [], ∀ m0. # x0 ≠ m0 + 1 ∨ x0[m0] = x0.last ⊦

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

spec: nat, name:elim-pred-c, seq:n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1, subst:[n, m] → [# x0, m]

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) ⊦ m + n = m + n0 ↔ n = n0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # x0, m

Proof Tree for at-last (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.