Proof step 1 from proof stutf-maxinf-is-fin (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  stutf(nσ) ⊦ max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m) = ∞ ↔ # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m) ∨ (∀ n. ⌜n⌝ < # nσ → nσ[n] < m)

Rule argument

Additional information

Proof step 2 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  stutf(nσ) ⊦ max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m) = ∞ ↔ # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m) ∨ (∀ n. ⌜n⌝ < # nσ → nσ[n] < m)

Rule argument

Additional information

Proof step 3 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))

• Proof step 5:

  stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ ⌜n0⌝ < # nσ ∧ nσ[n0] = m) ∨ (∀ n. ¬(⌜n⌝ < # nσ ∧ nσ[n] = m)) ↔ # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m) ∨ (∀ n. ⌜n⌝ < # nσ → nσ[n] < m)

Rule argument

Additional information

Used terms: λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m

Proof step 4 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Lemma maxinf-inf

Proof step 5 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ ⌜n0⌝ < # nσ ∧ nσ[n0] = m) ∨ (∀ n. ¬ ⌜n⌝ < # nσ ∨ nσ[n] ≠ m) ↔ # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m) ∨ (∀ n. ⌜n⌝ < # nσ → nσ[n] < m)

Rule argument

Additional information

Proof step 6 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 7:

  stutf(nσ), (∀ n. ∃ n0. n < n0 ∧ ⌜n0⌝ < # nσ ∧ nσ[n0] = m) ∨ (∀ n. ¬ ⌜n⌝ < # nσ ∨ nσ[n] ≠ m) ⊦ ∀ n. ⌜n⌝ < # nσ → nσ[n] < m, # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m)

• Proof step 34:

  stutf(nσ), # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m) ∨ (∀ n. ⌜n⌝ < # nσ → nσ[n] < m) ⊦ ∀ n. ∃ n0. n < n0 ∧ ⌜n0⌝ < # nσ ∧ nσ[n0] = m, ∀ n. ¬ ⌜n⌝ < # nσ ∨ nσ[n] ≠ m

Rule argument

Additional information

Proof step 7 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 8:

  stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ ⌜n0⌝ < # nσ ∧ nσ[n0] = m ⊦ ∀ n. ⌜n⌝ < # nσ → nσ[n] < m, # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m)

• Proof step 27:

  stutf(nσ), ∀ n. ¬ ⌜n⌝ < # nσ ∨ nσ[n] ≠ m ⊦ ∀ n. ⌜n⌝ < # nσ → nσ[n] < m, # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m)

Rule argument

Additional information

Proof step 8 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 9:

  # nσ = ∞, stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ ⌜n0⌝ < # nσ ∧ nσ[n0] = m ⊦ ∀ n. ⌜n⌝ < # nσ → nσ[n] < m, # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m)

• Proof step 22:

  # nσ ≠ ∞, stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ ⌜n0⌝ < # nσ ∧ nσ[n0] = m ⊦ ∀ n. ⌜n⌝ < # nσ → nσ[n] < m, # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m)

Rule argument

Additional information

Used formulas: # nσ = ∞

Proof step 9 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  # nσ = ∞, stutf(nσ), m ≤ nσ[n], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = m ⊦ ∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 10 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 11:

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

• Proof step 12:

  # nσ = ∞, stutf(nσ), m ≤ nσ[n], stutf(nσ) ∧ ⌜n⌝ < # nσ ∧ m ≤ nσ[n] → (∃ m0. m0 ≤ n ∧ nσ[m0] = m), ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = m ⊦ ∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m

Rule argument

Additional information

Used terms: nσ, n, m

Proof step 11 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 12 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  # nσ = ∞, stutf(nσ), m0 ≤ n, nσ[m0] ≤ nσ[n], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[m0] ⊦ ∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = nσ[m0]

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

Proof step 13 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 14:

  # nσ = ∞, stutf(nσ), nσ[m0] ≤ nσ[n], m0 ≤ n, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[m0] ⊦ ∀ n0. ¬ n0 < m0 → nσ[n0] = nσ[m0]

Rule argument

Additional information

Used terms: m0

Proof step 14 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  # nσ = ∞, nσ[n0] ≠ nσ[m0], stutf(nσ), m0 ≤ n, nσ[m0] ≤ nσ[n], m0 ≤ n0, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[m0] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 15 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 16:

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

• Proof step 17:

  # nσ = ∞, nσ[n0] ≠ nσ[m0], stutf(nσ), m0 ≤ n0, nσ[m0] ≤ nσ[n], m0 ≤ n, mon≤(nσ) ∧ m0 ≤ n0 ∧ ⌜n0⌝ < # nσ → nσ[m0] ≤ nσ[n0], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[m0] ⊦

Rule argument

Additional information

Used terms: nσ, m0, n0

Proof step 16 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Lemma monle-leq

Proof step 17 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  # nσ = ∞, nσ[n0] ≠ nσ[m0], nσ[m0] < nσ[n0], stutf(nσ), m0 ≤ n, nσ[m0] ≤ nσ[n], m0 ≤ n0, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[m0] ⊦

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 ⊦ ⌜m⌝ < ∞; (s) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)

Proof step 18 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 19:

  # nσ = ∞, nσ[n0] ≠ nσ[m0], nσ[m0] < nσ[n0], stutf(nσ), m0 ≤ n0, nσ[m0] ≤ nσ[n], m0 ≤ n, ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[m0] ⊦

Rule argument

Additional information

Used terms: n0

Proof step 19 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  # nσ = ∞, nσ[n1] = nσ[m0], nσ[n0] ≠ nσ[m0], m0 < n1, nσ[m0] < nσ[n0], n0 < n1, stutf(nσ), m0 ≤ n0, nσ[m0] ≤ nσ[n], m0 ≤ n ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat

Proof step 20 from proof stutf-maxinf-is-fin (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]

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → m ≤ n; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ m < n → ¬ n < m; (s) Used terms: nσ, n0, n1

Proof step 21 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Lemma monle-leq

Proof step 22 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  # nσ ≠ ∞, ⌜n⌝ < # nσ, stutf(nσ), m ≤ nσ[n], ∀ n. ∃ n0. n < n0 ∧ ⌜n0⌝ < # nσ ∧ nσ[n0] = m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 23 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 24:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 25:

  ⌜n⌝ < # nσ, stutf(nσ), m ≤ nσ[n], ∀ n. ∃ n0. n < n0 ∧ ⌜n0⌝ < # nσ ∧ nσ[n0] = m, ∃ n. # nσ = ⌜n⌝ ⊦

Rule argument

Additional information

Used terms: # nσ

Proof step 24 from proof stutf-maxinf-is-fin (previous step) (back to tree)

A spec-lemma

Proof step 25 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  # nσ = ⌜n0⌝ , n < n0, stutf(nσ), m ≤ nσ[n], ∀ n. ∃ n0. n < n0 ∧ ⌜n0⌝ < # nσ ∧ nσ[n0] = 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; (s)

Proof step 26 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

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

Proof step 27 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 28:

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

• Proof step 29:

  stutf(nσ), stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ m ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = m), ∀ n. ¬ ⌜n⌝ < # nσ ∨ nσ[n] ≠ m ⊦ ∀ n. ⌜n⌝ < # nσ → nσ[n] < m, # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m)

Rule argument

Additional information

Used terms: nσ, m, m

Proof step 28 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 29 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  ⌜n⌝ < # nσ, stutf(nσ), m ≤ nσ[n], ⌜m⌝ < # nσ ∧ m ≤ nσ[m] → m0 ≤ nσ[m0] ∧ nσ[m0] = m, ∀ n. ¬ ⌜n⌝ < # nσ ∨ nσ[n] ≠ m ⊦ # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 30 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 31:

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

• Proof step 32:

  ⌜n⌝ < # nσ, stutf(nσ), m ≤ nσ[n], ⌜m⌝ < # nσ ∧ m ≤ nσ[m] → m0 ≤ nσ[m0] ∧ nσ[m0] = m, stutf(nσ) ∧ ⌜n⌝ < # nσ ∧ m ≤ nσ[n] → (∃ m0. m0 ≤ n ∧ nσ[m0] = m), ∀ n. ¬ ⌜n⌝ < # nσ ∨ nσ[n] ≠ m ⊦ # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m)

Rule argument

Additional information

Used terms: nσ, n, m

Proof step 31 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 32 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  ⌜n⌝ < # nσ, stutf(nσ), m1 ≤ n, nσ[m1] ≤ nσ[n], ⌜nσ[m1]⌝ < # nσ ∧ nσ[m1] ≤ nσ[nσ[m1]] → m0 ≤ nσ[m0] ∧ nσ[m0] = nσ[m1], ∀ n. ¬ ⌜n⌝ < # nσ ∨ nσ[n] ≠ nσ[m1] ⊦ # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = nσ[m1])

Rule argument

Additional information

Proof step 33 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: m1

Proof step 34 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

  ⌜n0⌝ < # nσ, stutf(nσ), # nσ = ∞ ∧ (∀ n2. ¬ n2 < n1 → nσ[n2] = nσ[n0]) ∨ (∀ n. ⌜n⌝ < # nσ → nσ[n] < nσ[n0]) ⊦ ∃ n1. n < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = nσ[n0]

Rule argument

Additional information

Proof step 35 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 36:

  # nσ = ∞, ⌜n0⌝ < # nσ, stutf(nσ), ∀ n2. ¬ n2 < n1 → nσ[n2] = nσ[n0] ⊦ ∃ n1. n < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = nσ[n0]

• Proof step 44:

  ⌜n0⌝ < # nσ, stutf(nσ), ∀ n. ⌜n⌝ < # nσ → nσ[n] < nσ[n0] ⊦ ∃ n1. n < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = nσ[n0]

Rule argument

Additional information

Proof step 36 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  # nσ = ∞, stutf(nσ), ∀ n2. ¬ n2 < n1 → nσ[n2] = nσ[n0] ⊦ ∃ n2. n < n2 ∧ nσ[n2] = nσ[n0]

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

Proof step 37 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 38:

  # nσ = ∞, stutf(nσ), ¬ n1 < n1 → nσ[n1] = nσ[n0], ∀ n2. ¬ n2 < n1 → nσ[n2] = nσ[n0] ⊦ ∃ n2. n < n2 ∧ nσ[n2] = nσ[n0]

Rule argument

Additional information

Used terms: n1

Proof step 38 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

  # nσ = ∞, nσ[n1] = nσ[n0], stutf(nσ), ∀ n2. ¬ n2 < n1 → nσ[n2] = nσ[n0] ⊦ ∃ n2. n < n2 ∧ nσ[n2] = nσ[n0]

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)

Proof step 39 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 40:

  # nσ = ∞, nσ[n1] = nσ[n0], stutf(nσ), ∀ n2. ¬ n2 < n1 → nσ[n2] = nσ[n0] ⊦ ∃ n2. n < n2 ∧ nσ[n2] = nσ[n0], n < n1 ∧ nσ[n1] = nσ[n0]

Rule argument

Additional information

Used terms: n1

Proof step 40 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

  # nσ = ∞, nσ[n1] = nσ[n0], stutf(nσ), n1 ≤ n, ∀ n2. ¬ n2 < n1 → nσ[n2] = nσ[n0] ⊦ ∃ n2. n < n2 ∧ nσ[n2] = nσ[n0]

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 41 from proof stutf-maxinf-is-fin (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 42:

  # nσ = ∞, nσ[n1] = nσ[n0], stutf(nσ), n1 ≤ n, ∀ n2. ¬ n2 < n1 → nσ[n2] = nσ[n0] ⊦ ∃ n2. n < n2 ∧ nσ[n2] = nσ[n0], n < n + 1 ∧ nσ[n + 1] = nσ[n0]

Rule argument

Additional information

Used terms: n + 1

Proof step 42 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 43:

  # nσ = ∞, nσ[n1] = nσ[n0], nσ[n + 1] ≠ nσ[n0], stutf(nσ), n1 ≤ n, ∀ n2. ¬ n2 < n1 → nσ[n2] = nσ[n0] ⊦ ∃ n2. n < n2 ∧ nσ[n2] = nσ[n0]

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 ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s)

Proof step 43 from proof stutf-maxinf-is-fin (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m < n → n ≠ 0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: n + 1

Proof step 44 from proof stutf-maxinf-is-fin (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: n0