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Proof step 1 from proof leninv-inf (conclusion) (back to tree)

⊦ stutf(nσ) → (leninv(nσ) = ∞ ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]))

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof leninv-inf (previous step) (back to tree)

stutf(nσ) ⊦ leninv(nσ) = ∞ ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 3 from proof leninv-inf (previous step) (back to tree)

stutf(nσ) ⊦ leninv(nσ) = ∞ ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0])

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:leninv-def, seq:⊦ leninv(nσ) = max∞ (λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1 ∧ ¬(# nσ = ∞ ∧ (∀ m0. m < m0 → nσ[m0] = n - 1)))), subst:[nσ] → [nσ], <all paths>

Additional information

Used terms: nσ

Proof step 4 from proof leninv-inf (previous step) (back to tree)

⊦ leninv(nσ) = max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1 ∧ ¬(# nσ = ∞ ∧ (∀ m0. m < m0 → nσ[m0] = n - 1))))

Lemma leninv-def

Proof step 5 from proof leninv-inf (previous step) (back to tree)

stutf(nσ) ⊦ max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1 ∧ ¬(# nσ = ∞ ∧ (∀ m0. m < m0 → nσ[m0] = n - 1)))) = ∞ ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 6 from proof leninv-inf (previous step) (back to tree)

stutf(nσ) ⊦ max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n - 1)))) = ∞ ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0])

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:maxinf-inf, seq: ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n)), subst:[np] → [λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n - 1)))], <all paths>

Additional information

Used terms: λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n - 1)))

Proof step 7 from proof leninv-inf (previous step) (back to tree)

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

Lemma maxinf-inf

Proof step 8 from proof leninv-inf (previous step) (back to tree)

stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ (n0 ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1))))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n - 1))))) ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)

Proof step 9 from proof leninv-inf (previous step) (back to tree)

stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1)))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n - 1))))) ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0])

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: # nσ = ∞

Additional information

Used formulas: # nσ = ∞

Proof step 10 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1)))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n - 1))))) ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

Proof step 11 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ (∃ m. nσ[m] = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1))) ∨ (∀ n. ¬(n ≠ 0 → (∃ n0. nσ[n0] = n - 1 ∧ ¬(∀ m0. n0 < m0 → nσ[m0] = n - 1)))) ↔ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0])

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Proof step 12 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), (∀ n. ∃ n0. n < n0 ∧ (∃ m. nσ[m] = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1))) ∨ (∀ n. ¬(n ≠ 0 → (∃ n0. nσ[n0] = n - 1 ∧ ¬(∀ m0. n0 < m0 → nσ[m0] = n - 1)))) ⊦ ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 13 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), (∀ n. ∃ n0. n < n0 ∧ (∃ m. nσ[m] = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1))) ∨ (∀ n. ¬(n ≠ 0 → (∃ n0. nσ[n0] = n - 1 ∧ ¬(∀ m0. n0 < m0 → nσ[m0] = n - 1)))) ⊦ ∃ m0. n < m0 ∧ nσ[n] < nσ[m0]

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 3

Additional information

Proof step 14 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ m. nσ[m] = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1)) ⊦ ∃ m0. n < m0 ∧ nσ[n] < nσ[m0]

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 3 terms: nσ[n] + 1 abort: F discard: T computed: F

Additional information

Used terms: nσ[n] + 1

Proof step 15 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), ∃ n0. nσ[n] + 1 < n0 ∧ (∃ m. nσ[m] = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1)) ⊦ ∃ m0. n < m0 ∧ nσ[n] < nσ[m0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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 leninv-inf (previous step) (back to tree)

# nσ = ∞, nσ[n1] = n0 - 1, nσ[n2] ≠ n0 - 1, nσ[n] + 1 < n0, n1 < n2, stutf(nσ) ⊦ ∃ m0. n < m0 ∧ nσ[n] < nσ[m0]

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: nat, name:elim-pred-c, seq:n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1, subst:[n, m] → [n0, n3]

Additional information

Used simplifier rules: n0 < n ⊦ ¬ m + (* n) < (* n0); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n0, m

Proof step 17 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, nσ[n1] = n3, n0 = n3 + 1, nσ[n2] ≠ n3, n0 ≠ 0, n1 < n2, nσ[n] + 1 < n0, stutf(nσ) ⊦ ∃ m0. n < m0 ∧ nσ[n] < nσ[m0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 18 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, nσ[n2] ≠ nσ[n1], nσ[n] < nσ[n1], n1 < n2, stutf(nσ) ⊦ ∃ m0. n < m0 ∧ nσ[n] < nσ[m0]

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: n1 abort: F discard: T computed: T

Additional information

Used terms: n1

Proof step 19 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, nσ[n2] ≠ nσ[n1], n1 < n2, nσ[n] < nσ[n1], stutf(nσ) ⊦ n < n1 ∧ nσ[n] < nσ[n1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 20 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, nσ[n2] ≠ nσ[n1], nσ[n] < nσ[n1], n1 < n2, stutf(nσ), n1 ≤ n ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:monle-leq, seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n], subst:[nσ, m, n] → [nσ, n1, n]

Additional information

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

Proof step 21 from proof leninv-inf (previous step) (back to tree)

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

Lemma monle-leq

Proof step 22 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), ∀ n. ¬(n ≠ 0 → (∃ n0. nσ[n0] = n - 1 ∧ ¬(∀ m0. n0 < m0 → nσ[m0] = n - 1))) ⊦ ∃ m0. n < m0 ∧ nσ[n] < nσ[m0]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: 0

Additional information

Used terms: 0

Proof step 23 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ ∀ n. ∃ n0. n < n0 ∧ (∃ m. nσ[m] = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1)), ∀ n. ¬(n ≠ 0 → (∃ n0. nσ[n0] = n - 1 ∧ ¬(∀ m0. n0 < m0 → nσ[m0] = n - 1)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 24 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ ∃ n0. n < n0 ∧ (∃ m. nσ[m] = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1))

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: ∀ n. ∃ m. nσ[m] = n

Additional information

Used formulas: ∀ n. ∃ m. nσ[m] = n

Proof step 25 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n. ∃ m. nσ[m] = n ⊦ ∃ n0. n < n0 ∧ (∃ m. nσ[m] = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1))

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: n + 1 abort: F discard: T computed: T

Additional information

Used terms: n + 1

Proof step 26 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ n < n + 1 ∧ (∃ m. nσ[m] = n + 1 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n + 1 - 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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 + 0 = n; (s) ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s)

Proof step 27 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n. ∃ m. nσ[m] = n ⊦ ∃ m. nσ[m] = n ∧ ¬(∀ m0. m < m0 → nσ[m0] = nσ[m])

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 5 terms: n abort: F discard: F computed: F

Additional information

Used terms: n

Proof step 28 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∃ m. nσ[m] = n ⊦ ∃ m. nσ[m] = n ∧ ¬(∀ m0. m < m0 → nσ[m0] = nσ[m])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 29 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n. ∃ m. nσ[m] = n ⊦ ∃ m. nσ[m] = nσ[n3] ∧ ¬(∀ m0. m < m0 → nσ[m0] = nσ[m])

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: n3 abort: F discard: T computed: T

Additional information

Used terms: n3

Proof step 30 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[n3] = nσ[n3] ∧ ¬(∀ m0. n3 < m0 → nσ[m0] = nσ[n3])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 31 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n. ∃ m. nσ[m] = n, ∀ m0. n3 < m0 → nσ[m0] = nσ[n3] ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 5 terms: nσ[n3] + 1 abort: F discard: F computed: F

Additional information

Used terms: nσ[n3] + 1

Proof step 32 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ m0. n3 < m0 → nσ[m0] = nσ[n3], ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∃ m. nσ[m] = nσ[n3] + 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 33 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, nσ[n] = nσ[n3] + 1, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n. ∃ m. nσ[m] = n, ∀ m0. n3 < m0 → nσ[m0] = nσ[n3] ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 7 terms: n abort: F discard: T computed: F

Additional information

Used terms: n

Proof step 34 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, nσ[n] = nσ[n3] + 1, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), n3 < n → nσ[n] = nσ[n3], ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 35 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, nσ[n] = nσ[n3] + 1, stutf(nσ), n3 < n → nσ[n] = nσ[n3], n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n. ∃ m. nσ[m] = n ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:monle-leq, seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n], subst:[nσ, m, n] → [nσ, n, n3]

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m = n + 1 → ¬ m ≤ n; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ 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 Used terms: nσ, n, n3

Proof step 36 from proof leninv-inf (previous step) (back to tree)

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

Lemma monle-leq

Proof step 37 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ ∀ n. ∃ m. nσ[m] = n, ∃ n0. n < n0 ∧ (∃ m. nσ[m] = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 38 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), n0 ≠ 0 → nσ[n1] = n0 - 1 ∧ ¬(n1 < n2 → nσ[n2] = n0 - 1), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ ∃ n0. n < n0 ∧ (∃ m. nσ[m] = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1)), ∃ m. nσ[m] = n3

User applied rule weakening

And obtained the following new goals

Rule argument

poslist: left 3, right 1

Additional information

Proof step 39 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ ∃ m. nσ[m] = n3

User applied rule structural induction

And obtained the following new goals

Rule argument

indvar: n3

Additional information

Used simplifier rules: nat freely generated by 0, +1;;

Proof step 40 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ ∃ m. nσ[m] = 0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: 0 abort: F discard: T computed: F

Additional information

Used simplifier rules: stutf-zero: ⊦ stutf(nσ) → nσ[0] = 0; Used terms: 0

Proof step 41 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ nσ. # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → (∃ m. nσ[m] = n) ⊦ ∃ m. nσ[m] = n +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)

Proof step 42 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), ∀ nσ. # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → (∃ m. nσ[m] = n), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ ∃ m. nσ[m] = n + 1

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 3 terms: nσ abort: F discard: T computed: T

Additional information

Used terms: nσ

Proof step 43 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → (∃ m. nσ[m] = n), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ ∃ m. nσ[m] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 44 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ ∃ m. nσ[m] = nσ[n0] + 1

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 3 terms: n0 abort: F discard: T computed: F

Additional information

Used terms: n0

Proof step 45 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, stutf(nσ), ∃ m0. n0 < m0 ∧ nσ[n0] < nσ[m0] ⊦ ∃ m. nσ[m] = nσ[n0] + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 46 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, nσ[n0] < nσ[n], n0 < n, stutf(nσ) ⊦ ∃ m. nσ[m] = nσ[n0] + 1

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:stutf-alltaken, seq: ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n), subst:[nσ, m, n] → [nσ, n, nσ[n0] + 1]

Additional information

Used terms: nσ, n, nσ[n0] + 1

Proof step 47 from proof leninv-inf (previous step) (back to tree)

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

Lemma stutf-alltaken

Proof step 48 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, n0 < n, nσ[n0] < nσ[n], stutf(nσ), stutf(nσ) ∧ ⌜n⌝ < # nσ ∧ nσ[n0] + 1 ≤ nσ[n] → (∃ m0. m0 ≤ n ∧ nσ[m0] = nσ[n0] + 1) ⊦ ∃ m. nσ[m] = nσ[n0] + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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 eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ ⌜m⌝ < ∞; (s) ⊦ m + 1 ≤ n ↔ m < n; (s)

Proof step 49 from proof leninv-inf (previous step) (back to tree)

# nσ = ∞, nσ[n1] = nσ[n0] + 1, n0 < n, nσ[n0] < nσ[n], stutf(nσ), n1 ≤ n, nσ[n1] ≤ nσ[n] ⊦ ∃ m. nσ[m] = nσ[n1]

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: n1

Additional information

Used terms: n1

Proof step 50 from proof leninv-inf (previous step) (back to tree)

# nσ ≠ ∞, stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1)))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n - 1))))) ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 51 from proof leninv-inf (previous step) (back to tree)

# nσ ≠ ∞, stutf(nσ), (∀ n. ∃ n0. n < n0 ∧ (∃ n1. ⌜n1⌝ < # nσ ∧ nσ[n1] = n0 - 1)) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1))) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-inf, name:notinf-fin, seq: ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ ), subst:[n∞] → [# nσ], <all paths>

Additional information

Used terms: # nσ

Proof step 52 from proof leninv-inf (previous step) (back to tree)

⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

A spec-lemma

Proof step 53 from proof leninv-inf (previous step) (back to tree)

stutf(nσ), ∃ n. # nσ = ⌜n⌝ , (∀ n. ∃ n0. n < n0 ∧ (∃ n1. ⌜n1⌝ < # nσ ∧ nσ[n1] = n0 - 1)) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1))) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 54 from proof leninv-inf (previous step) (back to tree)

# nσ = ⌜n⌝ , stutf(nσ), (∀ n. ∃ n0. n < n0 ∧ (∃ n1. ⌜n1⌝ < # nσ ∧ nσ[n1] = n0 - 1)) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1))) ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 3

Additional information

Proof step 55 from proof leninv-inf (previous step) (back to tree)

# nσ = ⌜n⌝ , stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ n1. ⌜n1⌝ < # nσ ∧ nσ[n1] = n0 - 1) ⊦

User applied rule constructor cut

And obtained the following new goals

Rule argument

var: n

Additional information

Used simplifier rules: stutf-nonempty: ⊦ stutf(nσ) → (nσ = [] ↔ false); nat freely generated by 0, +1;; ⊦ # σ = ⌜0⌝ ↔ σ = []; (s)

Proof step 56 from proof leninv-inf (previous step) (back to tree)

# nσ = ⌜n⌝ , n = n2 +1, stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ n1. ⌜n1⌝ < # nσ ∧ nσ[n1] = n0 - 1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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) ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) 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 ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ m +1 = m + 1; (s)

Proof step 57 from proof leninv-inf (previous step) (back to tree)

# nσ = ⌜n2⌝ + ⌜1⌝ , stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ n1. ¬ n2 < n1 ∧ nσ[n1] = n0 - 1) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 3 terms: nσ[n2] + 1 abort: F discard: F computed: F

Additional information

Used terms: nσ[n2] + 1

Proof step 58 from proof leninv-inf (previous step) (back to tree)

# nσ = ⌜n2⌝ + ⌜1⌝ , stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ n1. ¬ n2 < n1 ∧ nσ[n1] = n0 - 1), ∃ n0. nσ[n2] + 1 < n0 ∧ (∃ n1. ¬ n2 < n1 ∧ nσ[n1] = n0 - 1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 59 from proof leninv-inf (previous step) (back to tree)

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: nat, name:elim-pred-c, seq:n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1, subst:[n, m] → [n0, n3]

Additional information

Used simplifier rules: n0 < n ⊦ ¬ m + (* n) < (* n0); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n0, m

Proof step 60 from proof leninv-inf (previous step) (back to tree)

# nσ = ⌜n2⌝ + ⌜1⌝ , nσ[n1] = n3, n0 = n3 + 1, n0 ≠ 0, nσ[n2] + 1 < n0, stutf(nσ), n1 ≤ n2, ∀ n. ∃ n0. n < n0 ∧ (∃ n1. ¬ n2 < n1 ∧ nσ[n1] = n0 - 1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 61 from proof leninv-inf (previous step) (back to tree)

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

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:monle-leq, seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n], subst:[nσ, m, n] → [nσ, n1, n2]

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ n < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: nσ, n1, n2

Proof step 62 from proof leninv-inf (previous step) (back to tree)

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

Lemma monle-leq

Proof step 63 from proof leninv-inf (previous step) (back to tree)

# nσ = ⌜n⌝ , stutf(nσ), ∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1)) ⊦

User applied rule constructor cut

And obtained the following new goals

Rule argument

var: n

Additional information

Used simplifier rules: stutf-nonempty: ⊦ stutf(nσ) → (nσ = [] ↔ false); nat freely generated by 0, +1;; ⊦ # σ = ⌜0⌝ ↔ σ = []; (s)

Proof step 64 from proof leninv-inf (previous step) (back to tree)

# nσ = ⌜n⌝ , n = n0 +1, stutf(nσ), ∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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) ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) 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 ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ m +1 = m + 1; (s)

Proof step 65 from proof leninv-inf (previous step) (back to tree)

# nσ = ⌜n0⌝ + ⌜1⌝ , stutf(nσ), ∀ n. ¬(n ≠ 0 → (∃ m. ¬ n0 < m ∧ nσ[m] = n - 1)) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 3 terms: nσ[n0] + 1 abort: F discard: F computed: F

Additional information

Used terms: nσ[n0] + 1

Proof step 66 from proof leninv-inf (previous step) (back to tree)

# nσ = ⌜n0⌝ + ⌜1⌝ , nσ[n0] + 1 ≠ 0, stutf(nσ), ∀ n. ¬(n ≠ 0 → (∃ m. ¬ n0 < m ∧ nσ[m] = n - 1)) ⊦ ∃ m. ¬ n0 < m ∧ nσ[m] = nσ[n0] + 1 - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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 + 0 = n; (s) ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 67 from proof leninv-inf (previous step) (back to tree)

# nσ = ⌜n0⌝ + ⌜1⌝ , stutf(nσ), ∀ n. ¬(n ≠ 0 → (∃ m. ¬ n0 < m ∧ nσ[m] = n - 1)) ⊦ ∃ m. ¬ n0 < m ∧ nσ[m] = nσ[n0]

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: n0

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s) Used terms: n0

Proof Tree for leninv-inf (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.

the proof tree