Proof step 1 from proof mininv-stays-if-not-taken (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  min(nσ, n + 1) ≠ min(nσ, n), mon<(nσ), ∀ m. ⌜m⌝ < # nσ → nσ(m) ≠ n ⊦

Rule argument

Additional information

Proof step 2 from proof mininv-stays-if-not-taken (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 3:

  ⊦ 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 4:

  min(nσ, n + 1) ≠ min(nσ, n), mon<(nσ), ∀ m. ⌜m⌝ < # nσ → nσ(m) ≠ n, 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) ) ) ⊦

Rule argument

Additional information

Used terms: nσ, n

Proof step 3 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Lemma min-def

Proof step 4 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  min(nσ, n + 1) ≠ min(nσ, n), nσ ≠ [], mon<(nσ), (# 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) ), ∀ m. ⌜m⌝ < # nσ → nσ(m) ≠ n ⊦

Rule argument

Additional information

Used simplifier rules: min-empty: ⊦ min([], n) = 0;

Proof step 5 from proof mininv-stays-if-not-taken (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 6:

  ⊦ 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 7:

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

Rule argument

Additional information

Used terms: nσ, n + 1

Proof step 6 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Lemma min-def

Proof step 7 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

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

Rule argument

Additional information

Used simplifier rules: 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 + 1 ↔ ¬ n < m; (s)

Proof step 8 from proof mininv-stays-if-not-taken (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 9:

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

• Proof step 15:

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

Rule argument

Additional information

Proof step 9 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

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

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m; (s) let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 10 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 11:

  # nσ = ⌜n0⌝ , min(nσ, n + 1) = n0, nσ(min(nσ, n + 1) - 1) = n, # nσ ≠ ∞, min(nσ, n + 1) ≠ min(nσ, n), nσ ≠ [], ⌜min(nσ, n)⌝ < # nσ, mon<(nσ), n ≤ nσ(min(nσ, n)), ∀ m. ⌜m⌝ < # nσ → nσ(m) ≠ n, ∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ(m) → min(nσ, n) ≤ m ⊦

Rule argument

Additional information

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

Proof step 11 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  # nσ = ⌜min(nσ, n + 1)⌝ , nσ(min(nσ, n + 1) - 1) = n, min(nσ, n + 1) ≠ min(nσ, n), nσ ≠ [], min(nσ, n) < min(nσ, n + 1), mon<(nσ), n ≤ nσ(min(nσ, n)), ∀ m. m < min(nσ, n + 1) ∧ n ≤ nσ(m) → min(nσ, n) ≤ m, ∀ m. m < min(nσ, nσ(m) + 1) → nσ(m) ≠ 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 12 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 13:

  # nσ = ⌜min(nσ, n + 1)⌝ , min(nσ, n + 1) = m0 + 1, nσ(m0) = n, min(nσ, n + 1) ≠ 0, min(nσ, n + 1) ≠ min(nσ, n), nσ ≠ [], min(nσ, n) < min(nσ, n + 1), mon<(nσ), n ≤ nσ(min(nσ, n)), ∀ m. m < min(nσ, nσ(m) + 1) → nσ(m) ≠ n, ∀ m. m < min(nσ, n + 1) ∧ n ≤ nσ(m) → min(nσ, n) ≤ m ⊦

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, n + 1), m0

Proof step 13 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  # nσ = ⌜m0⌝ + ⌜1⌝ , min(nσ, nσ(m0) + 1) = m0 + 1, m0 + 1 ≠ min(nσ, nσ(m0)), nσ ≠ [], min(nσ, nσ(m0)) ≤ m0, mon<(nσ), nσ(m0) ≤ nσ(min(nσ, nσ(m0))), ∀ m. m < min(nσ, nσ(m) + 1) → nσ(m) ≠ nσ(m0), ∀ m. ¬ m0 < m ∧ nσ(m0) ≤ nσ(m) → min(nσ, nσ(m0)) ≤ m ⊦

Rule argument

Additional information

Used simplifier rules: 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 < n ↔ ¬ n ≤ m; (ss) ⊦ 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 14 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

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 → ¬ m ≤ n; (s) Used terms: m0

Proof step 15 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

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

Rule argument

Additional information

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

Proof step 16 from proof mininv-stays-if-not-taken (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 17:

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m; (s)

Proof step 17 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

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

Rule argument

Additional information

Proof step 18 from proof mininv-stays-if-not-taken (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 19:

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

Rule argument

Additional information

Used terms: min(nσ, n + 1)

Proof step 19 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

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

Proof step 20 from proof mininv-stays-if-not-taken (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 21:

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

Rule argument

Additional information

Used terms: min(nσ, n)

Proof step 21 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

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

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m; (s) lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → ¬ n < m; (s) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)

Proof step 22 from proof mininv-stays-if-not-taken (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: min(nσ, n)