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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  min(nσ, n + 1) = 1 + min(nσ, n), nσ[min(nσ, n)] ≠ n, mon<(nσ) ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  min(nσ, n + 1) = 1 + min(nσ, n), nσ[min(nσ, n)] ≠ n, mon<(nσ) ⊦

Rule argument

Additional information

Proof step 3 from proof mininv-inverts (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, n + 1) = 1 + min(nσ, n), nσ[min(nσ, n)] ≠ 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) ) ) ⊦

Rule argument

Additional information

Used terms: nσ, n

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

Lemma min-def

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  min(nσ, n + 1) = 1 + min(nσ, n), nσ[min(nσ, n)] ≠ 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) ) ⊦

Rule argument

Additional information

Used simplifier rules: min-empty: ⊦ min([], n) = 0; c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 7:

  ⊦ 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 8:

  min(nσ, n + 1) = 1 + min(nσ, n), nσ[min(nσ, n)] ≠ 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) ) ⊦

Rule argument

Additional information

Used terms: nσ, n + 1

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

Lemma min-def

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

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

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

User applied rule case distinction

And obtained the following new goals

• Proof step 10:

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ n + 0 = 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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  min(nσ, n + 1) = 1 + min(nσ, n), nσ[min(nσ, n)] ≠ n, nσ ≠ [], ⌜1⌝ + ⌜min(nσ, n)⌝ < # nσ, n < nσ[1 + min(nσ, n)], 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 ∧ n < nσ[m] → min(nσ, n) < 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 11 from proof mininv-inverts (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 12:

  min(nσ, n + 1) = 1 + min(nσ, n), nσ[min(nσ, n)] ≠ n, nσ ≠ [], n < nσ[1 + min(nσ, n)], ⌜1⌝ + ⌜min(nσ, n)⌝ < # nσ, ⌜min(nσ, n)⌝ < # nσ, n < nσ[min(nσ, n)], mon<(nσ), ∀ m. ⌜m⌝ < # nσ ∧ n < nσ[m] → min(nσ, n) < m, ∀ 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 12 from proof mininv-inverts (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

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

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s) Used terms: min(nσ, n)