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Proof step 1 from proof min-stutters (conclusion) (back to tree)

mon<(nσ) ⊦ min(nσ, n + 1) = 1 + min(nσ, n) ∨ min(nσ, n + 1) = min(nσ, n)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof min-stutters (previous step) (back to tree)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 3 from proof min-stutters (previous step) (back to tree)

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

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:min-def, seq:⊦ 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) ) ), subst:[nσ, n] → [nσ, n]

Additional information

Used terms: nσ, n

Proof step 4 from proof min-stutters (previous step) (back to tree)

⊦ 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) ) )

Lemma min-def

Proof step 5 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ 1 + min(nσ, n), 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) ) ) ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:min-def, seq:⊦ 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) ) ), subst:[nσ, n] → [nσ, n + 1]

Additional information

Used terms: nσ, n + 1

Proof step 6 from proof min-stutters (previous step) (back to tree)

⊦ 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) ) )

Lemma min-def

Proof step 7 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ 1 + min(nσ, n), 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) ) ), 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) ) ) ⊦

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

Proof step 8 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ 1 + min(nσ, n), min(nσ, n + 1) ≠ min(nσ, n), mon<(nσ), (nσ = [] ⊃ min([], n + 1) = 0 ;(# 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) ) ), (nσ = [] ⊃ min([], 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) ) ) ⊦

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 4

Additional information

Proof step 9 from proof min-stutters (previous step) (back to tree)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 10 from proof min-stutters (previous step) (back to tree)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 5

Additional information

Proof step 11 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) = (# nσ).n, # nσ ≠ ∞, min(nσ, n + 1) ≠ 1 + min(nσ, n), 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) ) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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 12 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) = (# nσ).n, nσ(min(nσ, n + 1) - 1) = n, # nσ ≠ ∞, min(nσ, n + 1) ≠ 1 + min(nσ, n), 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 ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: nat-inf, name:elim-rqui, seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ), subst:[n∞, n] → [# nσ, n0]

Additional information

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

Proof step 13 from proof min-stutters (previous step) (back to tree)

# nσ = ⌜n0⌝ , min(nσ, n + 1) = n0, nσ(min(nσ, n + 1) - 1) = n, # nσ ≠ ∞, min(nσ, n + 1) ≠ 1 + min(nσ, n), 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 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 14 from proof min-stutters (previous step) (back to tree)

# nσ = ⌜min(nσ, n + 1)⌝ , nσ(min(nσ, n + 1) - 1) = n, min(nσ, n + 1) ≠ 1 + min(nσ, n), min(nσ, n + 1) ≠ min(nσ, n), nσ ≠ [], min(nσ, n) + 1 < min(nσ, n + 1), 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 ⊦

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] → [min(nσ, n + 1), m0]

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 15 from proof min-stutters (previous step) (back to tree)

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

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 → (m < n + 1 ↔ m < n); (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m + n = m + n0 ↔ n = n0; (s) ⊦ m + 1 ≠ 0; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic

Proof step 16 from proof min-stutters (previous step) (back to tree)

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

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:monls-ls, seq: ⊦ mon<(nσ) ∧ m < n ∧ ⌜n⌝ < # nσ → nσ(m) < nσ(n), subst:[nσ, m, n] → [nσ, min(nσ, nσ(m0)), m0]

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (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 ⊦ n∞0 + ⌜1⌝ = n∞ → (n∞0 < n∞ ↔ n∞0 ≠ ∞); (ss) ⊦ ⌜n⌝ ≠ ∞; (s) Used terms: nσ, min(nσ, nσ(m0)), m0

Proof step 17 from proof min-stutters (previous step) (back to tree)

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

Lemma monls-ls

Proof step 18 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ 1 + min(nσ, n), 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 ∧ n < nσ(m) → min(nσ, n + 1) ≤ m ⊦

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 19 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ min(nσ, n), min(nσ, n + 1) ≠ 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 ∧ n < nσ(m) → min(nσ, n + 1) ≤ m ⊦

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 8

Additional information

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

Proof step 20 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ min(nσ, n), min(nσ, n + 1) ≠ 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 ∧ n ≤ nσ(m) → min(nσ, n) ≤ m ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 21 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ min(nσ, n), min(nσ, n + 1) ≠ 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 ∧ n < nσ(m) → min(nσ, n + 1) ≤ m ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 11 terms: min(nσ, n + 1) abort: F discard: F computed: T

Additional information

Used terms: min(nσ, n + 1)

Proof step 22 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ min(nσ, n), min(nσ, n + 1) ≠ 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 ∧ 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) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ m < n → m ≤ n; (s)

Proof step 23 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ 1 + min(nσ, n), min(nσ, n + 1) ≠ min(nσ, n), nσ ≠ [], min(nσ, n) + 1 < min(nσ, n + 1), 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 ∧ n ≤ nσ(m) → min(nσ, n) ≤ m ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 13 terms: min(nσ, n) + 1 abort: F discard: F computed: T

Additional information

Used terms: min(nσ, n) + 1

Proof step 24 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ 1 + 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), min(nσ, n) + 1 < 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 ∧ n < nσ(m) → min(nσ, n + 1) ≤ m, ⌜min(nσ, n) + 1⌝ < # nσ ∧ n < nσ(min(nσ, n) + 1) → min(nσ, n + 1) ≤ min(nσ, 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 → (n ≤ m + 1 ↔ n = m + 1); (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 25 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ 1 + min(nσ, n), min(nσ, n + 1) ≠ min(nσ, n), nσ ≠ [], min(nσ, n) + 1 < min(nσ, n + 1), 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 ∧ n ≤ nσ(m) → min(nσ, n) ≤ m ⊦ ⌜min(nσ, n)⌝ + ⌜1⌝ < # nσ ∧ n < nσ(min(nσ, n) + 1)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Proof step 26 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ 1 + 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), min(nσ, n) + 1 < min(nσ, n + 1), mon<(nσ), n ≤ nσ(min(nσ, n)), n < nσ(min(nσ, n + 1)), ¬ ⌜min(nσ, n)⌝ + ⌜1⌝ < # nσ, # 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 ∧ n < nσ(m) → min(nσ, n + 1) ≤ m ⊦

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

Proof step 27 from proof min-stutters (previous step) (back to tree)

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

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 14 terms: min(nσ, n) abort: F discard: F computed: T

Additional information

Used terms: min(nσ, n)

Proof step 28 from proof min-stutters (previous step) (back to tree)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 29 from proof min-stutters (previous step) (back to tree)

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

User applied rule subst equation

And closed the goal!

Rule argument

pos: left 1, flags: rotate, discard

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s) n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s) ⊦ ⌜n⌝ .n = n; (s) 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 → m + n0 - n = (m - n) + n0; (s) ⊦ 0 + m = m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

Proof step 30 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ 1 + 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), min(nσ, n) + 1 < min(nσ, n + 1), mon<(nσ), n ≤ nσ(min(nσ, n)), n < nσ(min(nσ, n + 1)), ¬ 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 ∧ n < nσ(m) → min(nσ, n + 1) ≤ m ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 31 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ 1 + min(nσ, n), min(nσ, n + 1) ≠ min(nσ, n), nσ ≠ [], min(nσ, n) + 1 < min(nσ, n + 1), 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)), ¬ n < nσ(min(nσ, n) + 1), # 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 ∧ n ≤ nσ(m) → min(nσ, n) ≤ m ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:monls-ls, seq: ⊦ mon<(nσ) ∧ m < n ∧ ⌜n⌝ < # nσ → nσ(m) < nσ(n), subst:[nσ, m, n] → [nσ, min(nσ, n), min(nσ, n) + 1]

Additional information

Used terms: nσ, min(nσ, n), min(nσ, n) + 1

Proof step 32 from proof min-stutters (previous step) (back to tree)

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

Lemma monls-ls

Proof step 33 from proof min-stutters (previous step) (back to tree)

min(nσ, n + 1) ≠ 1 + 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), min(nσ, n) + 1 < min(nσ, n + 1), mon<(nσ), n ≤ nσ(min(nσ, n)), n < nσ(min(nσ, n + 1)), ¬ 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 ∧ n < nσ(m) → min(nσ, n + 1) ≤ m, mon<(nσ) ∧ min(nσ, n) < min(nσ, n) + 1 ∧ ⌜min(nσ, n) + 1⌝ < # nσ → nσ(min(nσ, n)) < nσ(min(nσ, n) + 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; (s) ⊦ n∞ = ⌜m⌝ + ⌜n⌝ → (n∞ = ∞ ↔ false); (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⌝ + ⌜1⌝ < n∞ ↔ ⌜m⌝ + ⌜1⌝ ≠ n∞); (ss) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s)

Proof step 34 from proof min-stutters (previous step) (back to tree)

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

User applied rule subst equation

And closed the goal!

Rule argument

pos: left 1, flags: rotate, discard

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s) n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s) ⊦ ⌜n⌝ .n = n; (s) ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ 0 + m = m; (s) ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; (s) ⊦ m ≠ n → (m < n + 1 ↔ m < n); (s) ⊦ m < n → ¬ n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic 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

Proof Tree for min-stutters (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