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

⊦ ⌜n⌝ < # σ → σ from n = σ[n] + (σ from n + 1)

System applied rule normalize

And obtained the following new goals

Proof step 2:
σ from n ≠ σ[n] + (σ from n + 1), ⌜n⌝ < # σ ⊦

Rule argument

Additional information

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

σ from n ≠ σ[n] + (σ from n + 1), ⌜n⌝ < # σ ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
σ from n ≠ ⌜σ[n] '⌝ + (σ from n + 1), ⌜n⌝ < # σ ⊦

Rule argument

Additional information

Used simplifier rules: cons-append: ⊦ a + σ = ⌜a '⌝ + σ;

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

σ from n ≠ ⌜σ[n] '⌝ + (σ from n + 1), ⌜n⌝ < # σ ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 4:
⌜x⌝ from n ≠ ⌜⌜x⌝ [n] '⌝ + (⌜x⌝ from n + 1), ⌜n⌝ < # ⌜x⌝ ⊦
Proof step 13:
⌜nf⌝ from n ≠ ⌜⌜nf⌝ [n] '⌝ + (⌜nf⌝ from n + 1), ⌜n⌝ < # ⌜nf⌝ ⊦

Rule argument

indvar: σ

Additional information

Used simplifier rules: stream('a) freely generated by ⌜ ⌝, ⌜ ⌝;;

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

⌜x⌝ from n ≠ ⌜⌜x⌝ [n] '⌝ + (⌜x⌝ from n + 1), ⌜n⌝ < # ⌜x⌝ ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
x[n] ' + (x from n + 1) ≠ x from n, x ≠ [], n < # x ⊦

Rule argument

Additional information

Used simplifier rules: sel-fin: ⊦ ⌜x⌝ [n] = x[n]; from-fin: ⊦ ⌜x⌝ from n = ⌜x from n⌝ ; eq-stream-to-list-2-1: ⊦ ⌜x⌝ + ⌜y⌝ = ⌜z⌝ ↔ x + y = z; append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ; ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)

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

x[n] ' + (x from n + 1) ≠ x from n, x ≠ [], n < # x ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 6:
(a + x0)[n] ' + (a + x0 from n + 1) ≠ a + x0 from n, a + x0 ≠ [], n < #(a + x0), ∀ n. ¬(x0[n] ' + (x0 from n + 1) ≠ x0 from n ∧ x0 ≠ [] ∧ n < # x0) ⊦

Rule argument

indvar: x

Additional information

Used simplifier rules: list('a) freely generated by [], +;;

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

(a + x0)[n] ' + (a + x0 from n + 1) ≠ a + x0 from n, a + x0 ≠ [], n < #(a + x0), ∀ n. ¬(x0[n] ' + (x0 from n + 1) ≠ x0 from n ∧ x0 ≠ [] ∧ n < # x0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
(a ' + x0)[n] ' + (x0 from n) ≠ a ' + x0 from n, n ≤ # x0, ∀ n0. x0[n0] ' + (x0 from n0 + 1) = x0 from n0 ∨ x0 = [] ∨ ¬ n0 < # x0 ⊦

Rule argument

Additional information

Used simplifier rules: from-succ-cons-l-lem: ⊦ a ' + x from n + 1 = x from n; ⊦ m < n ↔ ¬ n ≤ m; (ss) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s) ⊦ a + x = a ' + x; (s) ⊦ a ' + x ≠ []; (s)

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

(a ' + x0)[n] ' + (x0 from n) ≠ a ' + x0 from n, n ≤ # x0, ∀ n0. x0[n0] ' + (x0 from n0 + 1) = x0 from n0 ∨ x0 = [] ∨ ¬ n0 < # x0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 8:
n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]
Proof step 9:
x0[n -1] ' + (x0 from n) ≠ a ' + x0 from n, n ≠ 0, n ≤ # x0, ∀ n0. x0[n0] ' + (x0 from n0 + 1) = x0 from n0 ∨ x0 = [] ∨ ¬ n0 < # x0 ⊦

Rule argument

spec: list-del, name:get-numrec, seq:n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1], subst:[n, x, a] → [n, x0, a], <all paths>

Additional information

Used simplifier rules: from-zero-l: ⊦ x from 0 = x; ⊦ (a ' + x)[0] = a; (s) ⊦ # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0); (ws) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: n, x0, a

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

n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]

A spec-lemma

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

x0[n -1] ' + (x0 from n) ≠ a ' + x0 from n, n ≠ 0, n ≤ # x0, ∀ n0. x0[n0] ' + (x0 from n0 + 1) = x0 from n0 ∨ x0 = [] ∨ ¬ n0 < # x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
x0[n - 1] ' + (x0 from n) ≠ a ' + x0 from n, n ≠ 0, n ≤ # x0, ∀ n0. x0[n0] ' + (x0 from n0 + 1) = x0 from n0 ∨ x0 = [] ∨ ¬ n0 < # x0 ⊦

Rule argument

Additional information

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

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

x0[n - 1] ' + (x0 from n) ≠ a ' + x0 from n, n ≠ 0, n ≤ # x0, ∀ n0. x0[n0] ' + (x0 from n0 + 1) = x0 from n0 ∨ x0 = [] ∨ ¬ n0 < # x0 ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Proof step 11:
n = m + 1, x0[m] ' + (x0 from n) ≠ a ' + x0 from n, n ≠ 0, n ≤ # x0, ∀ n0. x0[n0] ' + (x0 from n0 + 1) = x0 from n0 ∨ x0 = [] ∨ ¬ n0 < # x0 ⊦

Rule argument

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

Additional information

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

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

n = m + 1, x0[m] ' + (x0 from n) ≠ a ' + x0 from n, n ≠ 0, n ≤ # x0, ∀ n0. x0[n0] ' + (x0 from n0 + 1) = x0 from n0 ∨ x0 = [] ∨ ¬ n0 < # x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
x0[m] ' + (x0 from m + 1) ≠ x0 from m, x0 ≠ [], m < # x0, ∀ n0. x0[n0] ' + (x0 from n0 + 1) = x0 from n0 ∨ ¬ n0 < # x0 ⊦

Rule argument

Additional information

Used simplifier rules: from-succ-cons-l-lem: ⊦ a ' + x from n + 1 = x from n; ⊦ # 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 ≠ 0; (s) a: ⊦ (x + y) + z = x + y + z; from specification list

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

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m

Additional information

Used terms: m

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

⌜nf⌝ from n ≠ ⌜⌜nf⌝ [n] '⌝ + (⌜nf⌝ from n + 1), ⌜n⌝ < # ⌜nf⌝ ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
⌜nf from n⌝ ≠ ⌜nf(n) '⌝ + ⌜nf from n + 1⌝ ⊦

Rule argument

Additional information

Used simplifier rules: sel-inf: ⊦ ⌜nf⌝ [n] = nf(n); from-inf: ⊦ ⌜nf⌝ from n = ⌜nf from n⌝ ; size-inf: ⊦ # ⌜nf⌝ = ∞; ⊦ ⌜m⌝ < ∞; (s)

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

⌜nf from n⌝ ≠ ⌜nf(n) '⌝ + ⌜nf from n + 1⌝ ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ nf from n = (λ m. nf(n + m))
Proof step 16:
⌜nf from n⌝ ≠ ⌜nf(n) '⌝ + ⌜λ m. nf((n + 1) + m)⌝ ⊦

Rule argument

spec: <toplevel>, name:from-nf, seq: ⊦ nf from n = (λ m. nf(n + m)), subst:[n, nf] → [n + 1, nf], <all paths>

Additional information

Used terms: n + 1, nf

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

⊦ nf from n = (λ m. nf(n + m))

Lemma from-nf

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

⌜nf from n⌝ ≠ ⌜nf(n) '⌝ + ⌜λ m. nf((n + 1) + m)⌝ ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ ⌜a '⌝ + ⌜nf⌝ = ⌜λ n. (n = 0 ⊃ a;nf(n -1))⌝
Proof step 18:
⌜nf from n⌝ ≠ ⌜λ n0. (n0 = 0 ⊃ nf(n);nf((n + 1) + n0 -1))⌝ ⊦

Rule argument

spec: <toplevel>, name:one-inf, seq: ⊦ ⌜a '⌝ + ⌜nf⌝ = ⌜λ n. (n = 0 ⊃ a;nf(n -1))⌝ , subst:[nf, a] → [λ m. nf((n + 1) + m), nf(n)], <all paths>

Additional information

Used terms: λ m. nf((n + 1) + m), nf(n)

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

⊦ ⌜a '⌝ + ⌜nf⌝ = ⌜λ n. (n = 0 ⊃ a;nf(n -1))⌝

Lemma one-inf

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

⌜nf from n⌝ ≠ ⌜λ n0. (n0 = 0 ⊃ nf(n);nf((n + 1) + n0 -1))⌝ ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
nf(n0 + n) ≠ (n0 = 0 ⊃ nf(n);nf(n + 1 + (n0 - 1))) ⊦

Rule argument

Additional information

Used simplifier rules: get-from-inf: ⊦ ((nf from n))(m) = nf(m + n); ⊦ ⌜nf⌝ = ⌜nf0⌝ ↔ nf = nf0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n -1 = n - 1; (s)

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

nf(n0 + n) ≠ (n0 = 0 ⊃ nf(n);nf(n + 1 + (n0 - 1))) ⊦

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 20:
n0 ≠ 0, nf(n0 + n) ≠ nf(n + 1 + (n0 - 1)) ⊦

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

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

n0 ≠ 0, nf(n0 + n) ≠ nf(n + 1 + (n0 - 1)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
n0 ≠ 0, nf(n0 + n) ≠ nf(n + 1 + (n0 - 1)) ⊦

Rule argument

Additional information

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

n0 ≠ 0, nf(n0 + n) ≠ nf(n + 1 + (n0 - 1)) ⊦

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] → [n0, m]

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0; (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: n0, m

Proof Tree for from-succ (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.