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

¬ # σ < ⌜n⌝ ⊦ m ≤ n → σ to n to m = σ to m

System applied rule normalize

And obtained the following new goals

Proof step 2:
σ to n to m ≠ σ to m, m ≤ n, ¬ # σ < ⌜n⌝ ⊦

Rule argument

Additional information

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

σ to n to m ≠ σ to m, m ≤ n, ¬ # σ < ⌜n⌝ ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
σ to n to m ≠ σ to m, m ≤ n, ¬ # σ < ⌜n⌝ ⊦

Rule argument

Additional information

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

σ to n to m ≠ σ to m, m ≤ n, ¬ # σ < ⌜n⌝ ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 4:
⌜x⌝ to n to m ≠ ⌜x⌝ to m, m ≤ n, ¬ # ⌜x⌝ < ⌜n⌝ ⊦

Rule argument

indvar: σ

Additional information

Used simplifier rules: to-fin: ⊦ ⌜x⌝ to n = ⌜x to n⌝ ; to-to-leq-inf: ⊦ m ≤ n → (nf to n) to m = nf to m; to-inf: ⊦ ⌜nf⌝ to n = ⌜nf to n⌝ ; stream freely generated by ⌜ ⌝, ⌜ ⌝;; ⊦ ⌜x⌝ = ⌜x0⌝ ↔ x = x0; (s)

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

⌜x⌝ to n to m ≠ ⌜x⌝ to m, m ≤ n, ¬ # ⌜x⌝ < ⌜n⌝ ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
x to n to m ≠ x to m, m ≤ n, n ≤ # x, m ≤ # x ⊦

Rule argument

Additional information

Used simplifier rules: to-fin: ⊦ ⌜x⌝ to n = ⌜x to n⌝ ; size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ; let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ⌜x⌝ = ⌜x0⌝ ↔ x = x0; (s)

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

x to n to m ≠ x to m, m ≤ n, n ≤ # x, m ≤ # x ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 6:
x to n0 +1 to m ≠ x to m, m ≤ # x, n0 +1 ≤ # x, m ≤ n0 +1, ∀ x, m. ¬(x to n0 to m ≠ x to m ∧ m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x) ⊦

Rule argument

indvar: n

Additional information

Used simplifier rules: to-zero-l: ⊦ x to 0 = []; nat freely generated by 0, +1;; ⊦ m ≤ 0 ↔ m = 0; (s) ⊦ 0 ≤ m; (s)

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

x to n0 +1 to m ≠ x to m, m ≤ # x, n0 +1 ≤ # x, m ≤ n0 +1, ∀ x, m. ¬(x to n0 to m ≠ x to m ∧ m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
x to n0 + 1 to m ≠ x to m, x ≠ [], n0 < # x, m ≤ n0 + 1, m ≤ # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ m +1 = m + 1; (s)

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

x to n0 + 1 to m ≠ x to m, x ≠ [], n0 < # x, m ≤ n0 + 1, m ≤ # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 8:
m < # x ⊦ x to m + 1 = (x to m) + x[m] '
Proof step 9:
(x to n0) + x[n0] ' to m ≠ x to m, x ≠ [], n0 < # x, m ≤ # x, m ≤ n0 + 1, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

spec: <toplevel>, name:to-unwind-end-fin, seq:m < # x ⊦ x to m + 1 = (x to m) + x[m] ', subst:[m, x] → [n0, x], <all paths>

Additional information

Used terms: n0, x

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

m < # x ⊦ x to m + 1 = (x to m) + x[m] '

Lemma to-unwind-end-fin

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

(x to n0) + x[n0] ' to m ≠ x to m, x ≠ [], n0 < # x, m ≤ # x, m ≤ n0 + 1, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ n ≤ # x → x + y to n = x to n
Proof step 11:
(x to n0) + x[n0] ' to m ≠ x to m, x ≠ [], n0 < # x, m ≤ n0 + 1, m ≤ # x, ¬ m ≤ #(x to n0), ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦
Proof step 25:
x to n0 to m ≠ x to m, x ≠ [], n0 < # x, m ≤ n0 + 1, m ≤ # x, m ≤ #(x to n0), ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

spec: <toplevel>, name:append-to-le, seq: ⊦ n ≤ # x → x + y to n = x to n, subst:[n, y, x] → [m, x[n0] ', x to n0], <all paths>

Additional information

Used terms: m, x[n0] ', x to n0

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

⊦ n ≤ # x → x + y to n = x to n

Lemma append-to-le

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

(x to n0) + x[n0] ' to m ≠ x to m, x ≠ [], n0 < # x, m ≤ n0 + 1, m ≤ # x, ¬ m ≤ #(x to n0), ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
(x to n0) + x[n0] ' to n0 + 1 ≠ x to n0 + 1, x ≠ [], n0 < # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n; 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 + 1 ≤ n ↔ m < n; (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) ⊦ m + 1 ≠ 0; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat 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) ⊦ m < n → ¬ n < m; (s)

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

(x to n0) + x[n0] ' to n0 + 1 ≠ x to n0 + 1, x ≠ [], n0 < # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 13:
m < # x ⊦ x to m + 1 = (x to m) + x[m] '
Proof step 14:
((x to n0) + x[n0] ' to n0) + ((x to n0) + x[n0] ')[n0] ' ≠ x to n0 + 1, x ≠ [], n0 < # x, n0 < #((x to n0) + x[n0] '), ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

spec: <toplevel>, name:to-unwind-end-fin, seq:m < # x ⊦ x to m + 1 = (x to m) + x[m] ', subst:[m, x] → [n0, (x to n0) + x[n0] '], <all paths>

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n; ⊦ #(x + y) = # x + # y; (s) ⊦ m < n → ¬ n < m; (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: n0, (x to n0) + x[n0] '

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

m < # x ⊦ x to m + 1 = (x to m) + x[m] '

Lemma to-unwind-end-fin

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

((x to n0) + x[n0] ' to n0) + ((x to n0) + x[n0] ')[n0] ' ≠ x to n0 + 1, x ≠ [], n0 < # x, n0 < #((x to n0) + x[n0] '), ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
((x to n0) + x[n0] ' to n0) + ((x to n0) + x[n0] ')[n0] ' ≠ x to n0 + 1, x ≠ [], n0 < # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

Additional information

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

((x to n0) + x[n0] ' to n0) + ((x to n0) + x[n0] ')[n0] ' ≠ x to n0 + 1, x ≠ [], n0 < # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ ¬ n < # x → (x + y)[n] = y[n - # x]
Proof step 17:
((x to n0) + x[n0] ' to n0) + x[n0] '[n0 - #(x to n0)] ' ≠ x to n0 + 1, x ≠ [], n0 < # x, ¬ n0 < #(x to n0), ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

spec: list-del, name:get-append-grt, seq: ⊦ ¬ n < # x → (x + y)[n] = y[n - # x], subst:[n, y, x] → [n0, x[n0] ', x to n0], <all paths>

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n; ⊦ n < # x → (x + y)[n] = x[n]; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m < n → ¬ n < m; (s) ⊦ ¬ n < n; (s) Used terms: n0, x[n0] ', x to n0

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

⊦ ¬ n < # x → (x + y)[n] = y[n - # x]

A spec-lemma

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

((x to n0) + x[n0] ' to n0) + x[n0] '[n0 - #(x to n0)] ' ≠ x to n0 + 1, x ≠ [], n0 < # x, ¬ n0 < #(x to n0), ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
((x to n0) + x[n0] ' to n0) + x[n0] ' ≠ x to n0 + 1, x ≠ [], n0 < # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n; ⊦ m ≤ m; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → ¬ n < m; (s) ⊦ a '[0] = a; (s) ⊦ m - m = 0; (s)

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

((x to n0) + x[n0] ' to n0) + x[n0] ' ≠ x to n0 + 1, x ≠ [], n0 < # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 19:
m < # x ⊦ x to m + 1 = (x to m) + x[m] '
Proof step 20:
((x to n0) + x[n0] ' to n0) + x[n0] ' ≠ (x to n0) + x[n0] ', x ≠ [], n0 < # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

spec: <toplevel>, name:to-unwind-end-fin, seq:m < # x ⊦ x to m + 1 = (x to m) + x[m] ', subst:[m, x] → [n0, x], <all paths>

Additional information

Used terms: n0, x

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

m < # x ⊦ x to m + 1 = (x to m) + x[m] '

Lemma to-unwind-end-fin

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

((x to n0) + x[n0] ' to n0) + x[n0] ' ≠ (x to n0) + x[n0] ', x ≠ [], n0 < # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
(x to n0) + x[n0] ' to n0 ≠ x to n0, x ≠ [], n0 < # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

Additional information

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

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

(x to n0) + x[n0] ' to n0 ≠ x to n0, x ≠ [], n0 < # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ n ≤ # x → x + y to n = x to n
Proof step 23:
x to n0 to n0 ≠ x to n0, x ≠ [], n0 < # x, n0 ≤ #(x to n0), ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

spec: <toplevel>, name:append-to-le, seq: ⊦ n ≤ # x → x + y to n = x to n, subst:[n, y, x] → [n0, x[n0] ', x to n0], <all paths>

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n; ⊦ m < n → ¬ n < m; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ ¬ n < n; (s) Used terms: n0, x[n0] ', x to n0

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

⊦ n ≤ # x → x + y to n = x to n

Lemma append-to-le

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

x to n0 to n0 ≠ x to n0, x ≠ [], n0 < # x, n0 ≤ #(x to n0), ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
x to n0 to n0 ≠ x to n0, x ≠ [], n0 < # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n; ⊦ m ≤ m; (s) ⊦ m < n → ¬ n < m; (s)

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

x to n0 to n0 ≠ x to n0, x ≠ [], n0 < # x, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: x, n0

Additional information

Used simplifier rules: ⊦ m ≤ m; (s) ⊦ m < n → m ≤ n; (s) Used terms: x, n0

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

x to n0 to m ≠ x to m, x ≠ [], n0 < # x, m ≤ n0 + 1, m ≤ # x, m ≤ #(x to n0), ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
x to n0 to m ≠ x to m, x ≠ [], m < # x, n0 < # x, m ≤ n0, m ≤ n0 + 1, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Rule argument

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n; ⊦ m < n → m ≤ n; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m < n → ¬ n < m; (s)

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

x to n0 to m ≠ x to m, x ≠ [], m < # x, n0 < # x, m ≤ n0, m ≤ n0 + 1, ∀ x0, m0. x0 to n0 to m0 = x0 to m0 ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # x0 ∨ ¬ m0 ≤ # x0 ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: x, m

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) Used terms: x, m

Proof Tree for to-to-leq (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.