Proof step 1 from proof mininv-inverts-inner (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  min(nσ, nσ(n)) ≠ n, mon<(nσ), ⌜n⌝ < # nσ ⊦

Rule argument

Additional information

Proof step 2 from proof mininv-inverts-inner (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  min(nσ, nσ(n)) ≠ n, mon<(nσ), ⌜n⌝ < # nσ ⊦

Rule argument

Additional information

Proof step 3 from proof mininv-inverts-inner (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 4:

  ⊦ mon<(nσ) → (nσ = [] ⊃ min(nσ, n) = 0 ;(# nσ ≠ ∞ ∧ nσ((# nσ).n - 1) < n ⊃ min(nσ, n) = (# nσ).n ;⌜min(nσ, n)⌝ < # nσ ∧ n ≤ nσ(min(nσ, n)) ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ(m) → min(nσ, n) ≤ m) ) )

• Proof step 5:

  min(nσ, nσ(n)) ≠ n, mon<(nσ), ⌜n⌝ < # nσ, mon<(nσ) → (nσ = [] ⊃ min(nσ, nσ(n)) = 0 ;(# nσ ≠ ∞ ∧ nσ((# nσ).n - 1) < nσ(n) ⊃ min(nσ, nσ(n)) = (# nσ).n ;⌜min(nσ, nσ(n))⌝ < # nσ ∧ nσ(n) ≤ nσ(min(nσ, nσ(n))) ∧ (∀ m. ⌜m⌝ < # nσ ∧ nσ(n) ≤ nσ(m) → min(nσ, nσ(n)) ≤ m) ) ) ⊦

Rule argument

Additional information

Used terms: nσ, nσ(n)

Proof step 4 from proof mininv-inverts-inner (previous step) (back to tree)

Lemma min-def

Proof step 5 from proof mininv-inverts-inner (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  min(nσ, nσ(n)) ≠ n, nσ ≠ [], mon<(nσ), ⌜n⌝ < # nσ, (# nσ ≠ ∞ ∧ nσ((# nσ).n - 1) < nσ(n) ⊃ min(nσ, nσ(n)) = (# nσ).n ;⌜min(nσ, nσ(n))⌝ < # nσ ∧ nσ(n) ≤ nσ(min(nσ, nσ(n))) ∧ (∀ m. ⌜m⌝ < # nσ ∧ nσ(n) ≤ nσ(m) → min(nσ, nσ(n)) ≤ m) ) ⊦

Rule argument

Additional information

Used simplifier rules: min-empty: ⊦ min([], n) = 0; ⊦ ¬ n < 0; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ # [] = ⌜0⌝ ; (s)

Proof step 6 from proof mininv-inverts-inner (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 7:

  min(nσ, nσ(n)) = (# nσ).n, # nσ ≠ ∞, min(nσ, nσ(n)) ≠ n, nσ ≠ [], mon<(nσ), ⌜n⌝ < # nσ, nσ((# nσ).n - 1) < nσ(n) ⊦

• Proof step 14:

  min(nσ, nσ(n)) ≠ n, nσ ≠ [], ⌜min(nσ, nσ(n))⌝ < # nσ, mon<(nσ), ⌜n⌝ < # nσ, nσ(n) ≤ nσ(min(nσ, nσ(n))), # nσ = ∞ ∨ ¬ nσ((# nσ).n - 1) < nσ(n), ∀ m. ⌜m⌝ < # nσ ∧ nσ(n) ≤ nσ(m) → min(nσ, nσ(n)) ≤ m ⊦

Rule argument

Additional information

Proof step 7 from proof mininv-inverts-inner (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  min(nσ, nσ(n)) = (# nσ).n, # nσ ≠ ∞, min(nσ, nσ(n)) ≠ n, nσ ≠ [], mon<(nσ), ⌜n⌝ < # nσ, nσ(min(nσ, nσ(n)) - 1) < nσ(n) ⊦

Rule argument

Additional information

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

Proof step 8 from proof mininv-inverts-inner (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 9:

  # nσ = ⌜n0⌝ , min(nσ, nσ(n)) = n0, # nσ ≠ ∞, min(nσ, nσ(n)) ≠ n, nσ ≠ [], mon<(nσ), ⌜n⌝ < # nσ, nσ(min(nσ, nσ(n)) - 1) < nσ(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: # nσ, n0

Proof step 9 from proof mininv-inverts-inner (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  # nσ = ⌜min(nσ, nσ(n))⌝ , min(nσ, nσ(n)) ≠ n, nσ ≠ [], mon<(nσ), n < min(nσ, nσ(n)), nσ(min(nσ, nσ(n)) - 1) < nσ(n) ⊦

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) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 10 from proof mininv-inverts-inner (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 11:

  # nσ = ⌜min(nσ, nσ(n))⌝ , min(nσ, nσ(n)) = n0 + 1, min(nσ, nσ(n)) ≠ 0, min(nσ, nσ(n)) ≠ n, nσ ≠ [], mon<(nσ), n < min(nσ, nσ(n)), nσ(n0) < nσ(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: min(nσ, nσ(n)), n0

Proof step 11 from proof mininv-inverts-inner (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  # nσ = ⌜n0⌝ + ⌜1⌝ , min(nσ, nσ(n)) = n0 + 1, n0 + 1 ≠ n, nσ ≠ [], mon<(nσ), n ≤ n0, nσ(n0) < nσ(n) ⊦

Rule argument

Additional information

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

Proof step 12 from proof mininv-inverts-inner (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 13:

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

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m < n → ¬ n < m; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) ⊦ m = m + n ↔ n = 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ¬ n < n; (s) Used terms: nσ, n, n0

Proof step 13 from proof mininv-inverts-inner (previous step) (back to tree)

Lemma monls-ls

Proof step 14 from proof mininv-inverts-inner (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  min(nσ, nσ(n)) ≠ n, nσ ≠ [], ⌜min(nσ, nσ(n))⌝ < # nσ, mon<(nσ), ⌜n⌝ < # nσ, nσ(n) ≤ nσ(min(nσ, nσ(n))), # nσ = ∞ ∨ ¬ nσ((# nσ).n - 1) < nσ(n), ∀ m. ⌜m⌝ < # nσ ∧ nσ(n) ≤ nσ(m) → min(nσ, nσ(n)) ≤ m ⊦

Rule argument

Additional information

Proof step 15 from proof mininv-inverts-inner (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 16:

  min(nσ, nσ(n)) ≠ n, nσ ≠ [], ⌜min(nσ, nσ(n))⌝ < # nσ, mon<(nσ), ⌜n⌝ < # nσ, nσ(n) ≤ nσ(min(nσ, nσ(n))), # nσ = ∞ ∨ ¬ nσ((# nσ).n - 1) < nσ(n), ∀ m. ⌜m⌝ < # nσ ∧ nσ(n) ≤ nσ(m) → min(nσ, nσ(n)) ≤ m, ⌜n⌝ < # nσ ∧ nσ(n) ≤ nσ(n) → min(nσ, nσ(n)) ≤ n ⊦

Rule argument

Additional information

Used terms: n

Proof step 16 from proof mininv-inverts-inner (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

  min(nσ, nσ(n)) ≠ n, nσ ≠ [], min(nσ, nσ(n)) < n, ⌜min(nσ, nσ(n))⌝ < # nσ, mon<(nσ), ⌜n⌝ < # nσ, nσ(n) ≤ nσ(min(nσ, nσ(n))), # nσ = ∞ ∨ ¬ nσ((# nσ).n - 1) < nσ(n), ∀ m. ⌜m⌝ < # nσ ∧ nσ(n) ≤ nσ(m) → min(nσ, nσ(n)) ≤ m ⊦

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 ↔ m < n); (s) ⊦ m ≤ m; (s)

Proof step 17 from proof mininv-inverts-inner (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 18:

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

Rule argument

Additional information

Used simplifier rules: mininv-inverts-weak: ⊦ mon<(nσ) ∧ ⌜min(nσ, n)⌝ < # nσ → (nσ(min(nσ, n)) < n ↔ false); Used terms: nσ, min(nσ, nσ(n)), n

Proof step 18 from proof mininv-inverts-inner (previous step) (back to tree)

Lemma monls-ls