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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(⌜n⌝ < leninv(nσ) ↔ true), ⌜m⌝ < # nσ, n < nσ[m], stutf(nσ) ⊦

Rule argument

Additional information

Proof step 2 from proof lsleninv-1 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ⌜m⌝ < # nσ, n < nσ[m], stutf(nσ), ¬ ⌜n⌝ < leninv(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 3 from proof lsleninv-1 (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 4:

  ⌜n0⌝ = leninv(nσ), n < nσ[m], ⌜m⌝ < # nσ, stutf(nσ), ¬ ⌜n⌝ < leninv(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: natinf freely generated by ∞, ⌜ ⌝;; ⊦ ⌜m⌝ < ∞; (s)

Proof step 4 from proof lsleninv-1 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  ⌜n0⌝ = leninv(nσ), n < nσ[m], ⌜m⌝ < # nσ, n0 < nσ[m], stutf(nσ), n0 ≤ n ⊦

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)

Proof step 5 from proof lsleninv-1 (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 6:

  n0 = 0, ⌜n0⌝ = leninv(nσ), n0 < nσ[m], ⌜m⌝ < # nσ, n < nσ[m], stutf(nσ), n0 ≤ n ⊦

• Proof step 10:

  n0 = n1 +1, ⌜n0⌝ = leninv(nσ), n0 < nσ[m], ⌜m⌝ < # nσ, n < nσ[m], stutf(nσ), n0 ≤ n ⊦

Rule argument

Additional information

Used simplifier rules: nat freely generated by 0, +1;;

Proof step 6 from proof lsleninv-1 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  ⌜0⌝ = leninv(nσ), n < nσ[m], ⌜m⌝ < # nσ, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) ⊦ 0 < m ↔ m ≠ 0; (s) ⊦ 0 ≤ m; (s)

Proof step 7 from proof lsleninv-1 (previous step) (back to tree)

User applied rule flip equation

And obtained the following new goals

• Proof step 8:

  leninv(nσ) = ⌜0⌝ , ⌜m⌝ < # nσ, n < nσ[m], stutf(nσ) ⊦

Rule argument

Additional information

Proof step 8 from proof lsleninv-1 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

  ⊦ stutf(nσ) → (leninv(nσ) = ⌜0⌝ ↔ nσ = ⌜λ n. 0⌝ )

Rule argument

Additional information

Used simplifier rules: ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ ⌜nf⌝ [n] = nf(n); (s) ⊦ ¬ n < 0; (s) Used terms: nσ

Proof step 9 from proof lsleninv-1 (previous step) (back to tree)

Lemma leninv-zero

Proof step 10 from proof lsleninv-1 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  ⌜n1⌝ + ⌜1⌝ = leninv(nσ), n1 < nσ[m], ⌜m⌝ < # nσ, n < nσ[m], n1 < n, n1 + 1 < nσ[m], stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic 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 + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 11 from proof lsleninv-1 (previous step) (back to tree)

User applied rule flip equation

And obtained the following new goals

• Proof step 12:

  leninv(nσ) = ⌜n1⌝ + ⌜1⌝ , n1 + 1 < nσ[m], n1 < n, n < nσ[m], ⌜m⌝ < # nσ, n1 < nσ[m], stutf(nσ) ⊦

Rule argument

Additional information

Proof step 12 from proof lsleninv-1 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

  ⊦ stutf(nσ) → (leninv(nσ) = ⌜n⌝ + ⌜1⌝ ↔ (# nσ = ∞ ⊃ ∃ m. nσ[m] = n ∧ (∀ m0. m < m0 → nσ[m0] = n + 1);nσ[(# nσ).n - 1] = n))

• Proof step 14:

  n1 < nσ[m], ⌜m⌝ < # nσ, n < nσ[m], n1 < n, n1 + 1 < nσ[m], stutf(nσ), (# nσ = ∞ ⊃ ∃ m. nσ[m] = n1 ∧ (∀ m0. m < m0 → nσ[m0] = n1 + 1);nσ[(# nσ).n - 1] = n1) ⊦

Rule argument

Additional information

Used terms: n1, nσ

Proof step 13 from proof lsleninv-1 (previous step) (back to tree)

Lemma leninv-succ

Proof step 14 from proof lsleninv-1 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  n1 + 1 < nσ[m], n1 < n, n < nσ[m], ⌜m⌝ < # nσ, n1 < nσ[m], stutf(nσ), (# nσ = ∞ ⊃ nσ[m0] = n1 ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1);nσ[(# nσ).n - 1] = n1) ⊦

Rule argument

Additional information

Proof step 15 from proof lsleninv-1 (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 16:

  ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n)

• Proof step 17:

  n1 < nσ[m], ⌜m⌝ < # nσ, n < nσ[m], n1 < n, n1 + 1 < nσ[m], stutf(nσ), stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n), (# nσ = ∞ ⊃ nσ[m0] = n1 ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1);nσ[(# nσ).n - 1] = n1) ⊦

Rule argument

Additional information

Used terms: nσ, m, n

Proof step 16 from proof lsleninv-1 (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 17 from proof lsleninv-1 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  n1 + 1 < nσ[m], n1 < nσ[m1], nσ[m1] < nσ[m], ⌜m⌝ < # nσ, n1 < nσ[m], stutf(nσ), m1 ≤ m, (# nσ = ∞ ⊃ nσ[m0] = n1 ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1);nσ[(# nσ).n - 1] = n1) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n → m ≤ n; (s)

Proof step 18 from proof lsleninv-1 (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 19:

  # nσ = ∞, nσ[m0] = n1, n1 < nσ[m], ⌜m⌝ < # nσ, nσ[m1] < nσ[m], n1 < nσ[m1], n1 + 1 < nσ[m], stutf(nσ), m1 ≤ m, ∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1 ⊦

• Proof step 24:

  nσ[(# nσ).n - 1] = n1, # nσ ≠ ∞, n1 < nσ[m], ⌜m⌝ < # nσ, nσ[m1] < nσ[m], n1 < nσ[m1], n1 + 1 < nσ[m], stutf(nσ), m1 ≤ m ⊦

Rule argument

Additional information

Proof step 19 from proof lsleninv-1 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  # nσ = ∞, nσ[m0] < nσ[m], nσ[m1] < nσ[m], nσ[m0] < nσ[m1], nσ[m0] + 1 < nσ[m], stutf(nσ), m1 ≤ m, ∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s) ⊦ m +1 = m + 1; (s) trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic

Proof step 20 from proof lsleninv-1 (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 21:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]

• Proof step 22:

  # nσ = ∞, nσ[m0] + 1 < nσ[m], nσ[m0] < nσ[m1], nσ[m1] < nσ[m], nσ[m0] < nσ[m], stutf(nσ), m1 ≤ m, mon≤(nσ) ∧ m1 ≤ m0 ∧ ⌜m0⌝ < # nσ → nσ[m1] ≤ nσ[m0], ∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1 ⊦

Rule argument

Additional information

Used terms: nσ, m1, m0

Proof step 21 from proof lsleninv-1 (previous step) (back to tree)

Lemma monle-leq

Proof step 22 from proof lsleninv-1 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  # nσ = ∞, nσ[m0] + 1 < nσ[m], nσ[m0] < nσ[m1], nσ[m1] < nσ[m], nσ[m0] < nσ[m], m0 < m1, m0 < m, stutf(nσ), m1 ≤ m, ∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1 ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ ⌜m⌝ < ∞; (s) ⊦ m < n → (n ≤ m ↔ false); (s)

Proof step 23 from proof lsleninv-1 (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: m

Proof step 24 from proof lsleninv-1 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  # nσ ≠ ∞, nσ[(# nσ).n - 1] < nσ[m], ⌜m⌝ < # nσ, nσ[m1] < nσ[m], nσ[(# nσ).n - 1] < nσ[m1], nσ[(# nσ).n - 1] + 1 < nσ[m], stutf(nσ), m1 ≤ m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s) trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic

Proof step 25 from proof lsleninv-1 (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 26:

  # nσ = ⌜n⌝ , # nσ ≠ ∞, nσ[n - 1] + 1 < nσ[m], nσ[n - 1] < nσ[m1], nσ[m1] < nσ[m], ⌜m⌝ < # nσ, nσ[n - 1] < nσ[m], stutf(nσ), m1 ≤ m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: # nσ, n

Proof step 26 from proof lsleninv-1 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  # nσ = ⌜n⌝ , m1 < n, m < n, nσ[m1] < nσ[m], nσ[n - 1] < nσ[m1], nσ[n - 1] < nσ[m], nσ[n - 1] + 1 < nσ[m], stutf(nσ), m1 ≤ m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ m +1 = m + 1; (s) ⊦ ⌜n⌝ ≠ ∞; (s) trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic

Proof step 27 from proof lsleninv-1 (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 28:

  # nσ = ⌜n⌝ , n = m0 + 1, n ≠ 0, nσ[m0] + 1 < nσ[m], nσ[m0] < nσ[m], nσ[m0] < nσ[m1], nσ[m1] < nσ[m], m < n, m1 < n, stutf(nσ), m1 ≤ m ⊦

Rule argument

Additional information

Used simplifier rules: stutf-nonempty: ⊦ stutf(nσ) → (nσ = [] ↔ false); ⊦ # σ = ⌜0⌝ ↔ σ = []; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0

Proof step 28 from proof lsleninv-1 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ[m0] < nσ[m], nσ[m0] < nσ[m1], nσ[m1] < nσ[m], nσ[m0] + 1 < nσ[m], stutf(nσ), m ≤ m0, m1 ≤ m, m1 ≤ m0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s) trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ 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) ⊦ m + 1 ≠ 0; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 29 from proof lsleninv-1 (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 30:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) ⊦ 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 ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: nσ, m1, m0

Proof step 30 from proof lsleninv-1 (previous step) (back to tree)

Lemma monle-leq