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), 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) ) ), ∀ m. ⌜m⌝ < # nσ → nσ[m] ≠ n ⊦

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

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:

  nσ[min(nσ, n + 1) - 1] = n, min(nσ, n + 1) = (# nσ).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⌝ , nσ[min(nσ, n + 1) - 1] = n, min(nσ, n + 1) = n0, # 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)⌝ , nσ[n0] = n, min(nσ, n + 1) = n0 + 1, 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 = ⌜n0⌝ + ⌜1⌝ , min(nσ, nσ[n0] + 1) = n0 + 1, n0 + 1 ≠ min(nσ, nσ[n0]), nσ ≠ [], mon<(nσ), nσ[n0] ≤ nσ[min(nσ, nσ[n0])], min(nσ, nσ[n0]) ≤ n0, ∀ m. m < min(nσ, nσ[m] + 1) → nσ[m] ≠ nσ[n0], ∀ m. ¬ n0 < m ∧ nσ[n0] ≤ nσ[m] → min(nσ, nσ[n0]) ≤ 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) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m = n + 1 → ¬ m ≤ n; (s) Used terms: n0

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

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

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, n < nσ[min(nσ, n + 1)], mon<(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, # nσ = ∞ ∨ ¬ nσ[(# nσ).n - 1] < n, # nσ = ∞ ∨ n < nσ[(# nσ).n - 1] ⊦

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

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

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, n < nσ[min(nσ, n + 1)], min(nσ, n) < min(nσ, n + 1), mon<(nσ), ∀ 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, # nσ = ∞ ∨ ¬ nσ[(# nσ).n - 1] < n, # nσ = ∞ ∨ n < nσ[(# nσ).n - 1] ⊦

Rule argument

Additional information

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