Proof step 1 from proof clean0-plus-clean1-bounded (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  #c mx ≠ 0, #r mx < #clean0rows(mx, i) + #clean1rows(mx, i) ⊦

Rule argument

Additional information

Proof step 2 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 3:

  ⊦ ¬ #r mx < #clean1rows(mx, i)

• Proof step 4:

  #c mx ≠ 0, #r mx < #clean0rows(mx, i) + #clean1rows(mx, i), ¬ #r mx < #clean1rows(mx, i) ⊦

Rule argument

Additional information

Used terms: mx, i

Proof step 3 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Lemma clean1rows-bounded

Proof step 4 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #r mx < #clean0rows(mx, i) + #clean1rows(mx, i), #clean1rows(mx, i) ≤ #r mx ⊦

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 < m + n ↔ n ≠ 0; (s) ⊦ 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 ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 5 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 6:

  ⊦ ¬ #r mx < #clean0rows(mx, i)

• Proof step 7:

  #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #r mx < #clean0rows(mx, i) + #clean1rows(mx, i), #clean1rows(mx, i) ≤ #r mx, ¬ #r mx < #clean0rows(mx, i) ⊦

Rule argument

Additional information

Used terms: mx, i

Proof step 6 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Lemma clean0rows-bounded

Proof step 7 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 0, #r mx < #clean0rows(mx, i) + #clean1rows(mx, i), #clean0rows(mx, i) ≤ #r mx, #clean1rows(mx, i) ≤ #r mx ⊦

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 < m + n ↔ n ≠ 0; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 8 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 9:

  #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 0, #r mx < #clean0rows(mx, i) + #clean1rows(mx, i), #clean0rows(mx, i) < #r mx, #clean1rows(mx, i) ≤ #r mx, #clean0rows(mx, i) ≤ #r mx ⊦

• Proof step 27:

  #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 0, #r mx < #clean0rows(mx, i) + #clean1rows(mx, i), #clean1rows(mx, i) ≤ #r mx, #clean0rows(mx, i) ≤ #r mx, ¬ #clean0rows(mx, i) < #r mx ⊦

Rule argument

Additional information

Used formulas: #clean0rows(mx, i) < #r mx

Proof step 9 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 1, #r mx < #clean0rows(mx, i) + #clean1rows(mx, i), #clean0rows(mx, i) < #r mx, #clean1rows(mx, i) ≤ #r mx ⊦

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 ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m +1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n → m ≤ n; (s)

Proof step 10 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 11:

  ⊦ r < #clean0rows(mx, i) → count(mx[r]r, λ j. ¬ j < i) = 0

• Proof step 12:

  #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 1, #clean0rows(mx, i) < #r mx, #r mx < #clean0rows(mx, i) + #clean1rows(mx, i), #clean1rows(mx, i) ≤ #r mx, #clean0rows(mx, i) - 1 < #clean0rows(mx, i) → count(mx[#clean0rows(mx, i) - 1]r, λ j. ¬ j < i) = 0 ⊦

Rule argument

Additional information

Used terms: mx, i, #clean0rows(mx, i) - 1

Proof step 11 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Lemma clean0rows-allls

Proof step 12 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  count(mx[#clean0rows(mx, i) - 1]r, λ j. ¬ j < i) = 0, #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 1, #r mx < #clean0rows(mx, i) + #clean1rows(mx, i), #clean0rows(mx, i) < #r mx, #clean1rows(mx, i) ≤ #r mx ⊦

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

Proof step 13 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 14:

  #clean0rows(mx, i) = m + 1, count(mx[m]r, λ j. ¬ j < i) = 0, #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 1, #clean0rows(mx, i) < #r mx, #r mx < #clean0rows(mx, i) + #clean1rows(mx, i), #clean1rows(mx, i) ≤ #r mx ⊦

Rule argument

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: #clean0rows(mx, i), m

Proof step 14 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  #clean0rows(mx, i) = m + 1, count(mx[m]r, λ j. ¬ j < i) = 0, #c mx ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 1, m + 1 < #r mx, #r mx ≤ m + #clean1rows(mx, i), #clean1rows(mx, i) ≤ #r mx ⊦

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 ≤ m + n; (s) ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) ⊦ m + n < m + n0 ↔ n < n0; (s) let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ m + n < m; (s) ⊦ m ≠ 1 → (m + n = 1 ↔ m = 0 ∧ n = 1); (ws) ⊦ m + 1 ≠ 0; (s) n0 < n ⊦ (* n) < m0 + (* n0) ↔ n - n0 < m0; (s) ⊦ m +1 = m + 1; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic trans-nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic

Proof step 15 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 16:

  ⊦ #r mx ≤ #clean1rows(mx, i) + r ∧ r < #r mx → count(mx[r]r, λ j. j < i) = 0

• Proof step 17:

  #clean0rows(mx, i) = m + 1, count(mx[m]r, λ j. ¬ j < i) = 0, #c mx ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 1, m + 1 < #r mx, #clean1rows(mx, i) ≤ #r mx, #r mx ≤ m + #clean1rows(mx, i), #r mx ≤ #clean1rows(mx, i) + m ∧ m < #r mx → count(mx[m]r, λ j. j < i) = 0 ⊦

Rule argument

Additional information

Used terms: mx, i, m

Proof step 16 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Lemma clean1rows-allge

Proof step 17 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  #clean0rows(mx, i) = m + 1, count(mx[m]r, λ j. ¬ j < i) = 0, count(mx[m]r, λ j. j < i) = 0, #c mx ≠ 0, #clean1rows(mx, i) ≠ 1, #clean1rows(mx, i) ≠ 0, m + 1 < #r mx, m < #r mx, #clean1rows(mx, i) ≤ #r mx, #r mx ≤ m + #clean1rows(mx, i) ⊦

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) ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ m < m + n ↔ n ≠ 0; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ ¬ m < n → ¬ m + m0 < n; (s)

Proof step 18 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

  ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n]))

• Proof step 20:

  #clean0rows(mx, i) = m + 1, count(mx[m]r, λ j. ¬ j < i) = 0, #c mx ≠ 0, #clean1rows(mx, i) ≠ 1, #clean1rows(mx, i) ≠ 0, m < #r mx, m + 1 < #r mx, #r mx ≤ m + #clean1rows(mx, i), #clean1rows(mx, i) ≤ #r mx, ∀ n. n < # mx[m]r → ¬ mx[m]r[n] < i ⊦

Rule argument

Additional information

Used terms: mx[m]r, λ j. j < i

Proof step 19 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

A spec-lemma

Proof step 20 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  #clean0rows(mx, i) = m + 1, count(mx[m]r, λ j. ¬ j < i) = 0, #c mx ≠ 0, #clean1rows(mx, i) ≠ 1, #clean1rows(mx, i) ≠ 0, m + 1 < #r mx, m < #r mx, #clean1rows(mx, i) ≤ #r mx, #r mx ≤ m + #clean1rows(mx, i), ∀ n. n < #c mx → ¬ mx[m, n] < i ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ mx[r]r[c] = mx[r, c]; (s) ⊦ # mx[r]r = #c mx; (s)

Proof step 21 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 22:

  ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n]))

• Proof step 23:

  #clean0rows(mx, i) = m + 1, #c mx ≠ 0, #clean1rows(mx, i) ≠ 1, #clean1rows(mx, i) ≠ 0, m < #r mx, m + 1 < #r mx, #r mx ≤ m + #clean1rows(mx, i), #clean1rows(mx, i) ≤ #r mx, ∀ n. n < #c mx → ¬ mx[m, n] < i, ∀ n. n < # mx[m]r → ¬ ¬ mx[m]r[n] < i ⊦

Rule argument

Additional information

Used terms: mx[m]r, λ j. ¬ j < i

Proof step 22 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

A spec-lemma

Proof step 23 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  #clean0rows(mx, i) = m + 1, #c mx ≠ 0, #clean1rows(mx, i) ≠ 1, #clean1rows(mx, i) ≠ 0, m + 1 < #r mx, m < #r mx, #clean1rows(mx, i) ≤ #r mx, #r mx ≤ m + #clean1rows(mx, i), ∀ n. n < #c mx → ¬ mx[m, n] < i, ∀ n. n < #c mx → mx[m, n] < i ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ mx[r]r[c] = mx[r, c]; (s) ⊦ # mx[r]r = #c mx; (s)

Proof step 24 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 25:

  #clean0rows(mx, i) = m + 1, #c mx ≠ 0, #clean1rows(mx, i) ≠ 1, #clean1rows(mx, i) ≠ 0, m < #r mx, m + 1 < #r mx, #r mx ≤ m + #clean1rows(mx, i), #clean1rows(mx, i) ≤ #r mx, 0 < #c mx → ¬ mx[m, 0] < i, ∀ n. n < #c mx → mx[m, n] < i ⊦

Rule argument

Additional information

Used terms: 0

Proof step 25 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  #clean0rows(mx, i) = m + 1, #c mx ≠ 0, #clean1rows(mx, i) ≠ 1, #clean1rows(mx, i) ≠ 0, m + 1 < #r mx, m < #r mx, #clean1rows(mx, i) ≤ #r mx, #r mx ≤ m + #clean1rows(mx, i), ¬ mx[m, 0] < i, ∀ n. n < #c mx → mx[m, n] < i ⊦

Rule argument

Additional information

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

Proof step 26 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < m ↔ m ≠ 0; (s) Used terms: 0

Proof step 27 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  #clean0rows(mx, i) = #r mx, #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≤ #clean0rows(mx, i) ⊦

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 < m + n ↔ n ≠ 0; (s) ⊦ m ≤ m; (s) let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (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)

Proof step 28 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 29:

  ⊦ #r mx ≤ #clean1rows(mx, i) + r ∧ r < #r mx → count(mx[r]r, λ j. j < i) = 0

• Proof step 30:

  #clean0rows(mx, i) = #r mx, #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≤ #clean0rows(mx, i), #r mx ≤ #clean1rows(mx, i) + (#r mx - 1) ∧ #r mx - 1 < #r mx → count(mx[#r mx - 1]r, λ j. j < i) = 0 ⊦

Rule argument

Additional information

Used terms: mx, i, #r mx - 1

Proof step 29 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Lemma clean1rows-allge

Proof step 30 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  #clean0rows(mx, i) = #r mx, #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≤ #clean0rows(mx, i), #clean0rows(mx, i) ≤ #clean1rows(mx, i) + (#clean0rows(mx, i) - 1) → count(mx[#clean0rows(mx, i) - 1]r, λ j. j < i) = 0 ⊦

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

Proof step 31 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 32:

  #clean0rows(mx, i) = #r mx, #clean0rows(mx, i) = m + 1, #c mx ≠ 0, #clean0rows(mx, i) ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≤ #clean0rows(mx, i), #clean0rows(mx, i) ≤ #clean1rows(mx, i) + m → count(mx[m]r, λ j. j < i) = 0 ⊦

Rule argument

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: #clean0rows(mx, i), m

Proof step 32 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  #clean0rows(mx, i) = m + 1, m + 1 = #r mx, count(mx[m]r, λ j. j < i) = 0, #c mx ≠ 0, #clean1rows(mx, i) ≠ 0, 1 ≤ #clean1rows(mx, i), #clean1rows(mx, i) ≤ m + 1 ⊦

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 ≤ m + n; (s) let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat ⊦ m < 1 ↔ m = 0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 33 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 34:

  ⊦ r < #clean0rows(mx, i) → count(mx[r]r, λ j. ¬ j < i) = 0

• Proof step 35:

  #clean0rows(mx, i) = m + 1, m + 1 = #r mx, count(mx[m]r, λ j. j < i) = 0, #c mx ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≤ m + 1, 1 ≤ #clean1rows(mx, i), m < #clean0rows(mx, i) → count(mx[m]r, λ j. ¬ j < i) = 0 ⊦

Rule argument

Additional information

Used terms: mx, i, m

Proof step 34 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Lemma clean0rows-allls

Proof step 35 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

  #clean0rows(mx, i) = m + 1, m + 1 = #r mx, count(mx[m]r, λ j. j < i) = 0, count(mx[m]r, λ j. ¬ j < i) = 0, #c mx ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≤ m + 1, 1 ≤ #clean1rows(mx, i) ⊦

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 36 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 37:

  ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n]))

• Proof step 38:

  #clean0rows(mx, i) = m + 1, m + 1 = #r mx, count(mx[m]r, λ j. j < i) = 0, #c mx ≠ 0, #clean1rows(mx, i) ≠ 0, 1 ≤ #clean1rows(mx, i), #clean1rows(mx, i) ≤ m + 1, ∀ n. n < # mx[m]r → ¬ ¬ mx[m]r[n] < i ⊦

Rule argument

Additional information

Used terms: mx[m]r, λ j. ¬ j < i

Proof step 37 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

A spec-lemma

Proof step 38 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

  #clean0rows(mx, i) = m + 1, m + 1 = #r mx, count(mx[m]r, λ j. j < i) = 0, #c mx ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≤ m + 1, 1 ≤ #clean1rows(mx, i), ∀ n. n < #c mx → mx[m, n] < i ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ mx[r]r[c] = mx[r, c]; (s) ⊦ # mx[r]r = #c mx; (s)

Proof step 39 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 40:

  ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n]))

• Proof step 41:

  #clean0rows(mx, i) = m + 1, m + 1 = #r mx, #c mx ≠ 0, #clean1rows(mx, i) ≠ 0, 1 ≤ #clean1rows(mx, i), #clean1rows(mx, i) ≤ m + 1, ∀ n. n < #c mx → mx[m, n] < i, ∀ n. n < # mx[m]r → ¬ mx[m]r[n] < i ⊦

Rule argument

Additional information

Used terms: mx[m]r, λ j. j < i

Proof step 40 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

A spec-lemma

Proof step 41 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  #clean0rows(mx, i) = m + 1, m + 1 = #r mx, #c mx ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≤ m + 1, 1 ≤ #clean1rows(mx, i), ∀ n. n < #c mx → mx[m, n] < i, ∀ n. n < #c mx → ¬ mx[m, n] < i ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ mx[r]r[c] = mx[r, c]; (s) ⊦ # mx[r]r = #c mx; (s)

Proof step 42 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 43:

  #clean0rows(mx, i) = m + 1, m + 1 = #r mx, #c mx ≠ 0, #clean1rows(mx, i) ≠ 0, 1 ≤ #clean1rows(mx, i), #clean1rows(mx, i) ≤ m + 1, 0 < #c mx → ¬ mx[m, 0] < i, ∀ n. n < #c mx → mx[m, n] < i ⊦

Rule argument

Additional information

Used terms: 0

Proof step 43 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

  #clean0rows(mx, i) = m + 1, m + 1 = #r mx, #c mx ≠ 0, #clean1rows(mx, i) ≠ 0, #clean1rows(mx, i) ≤ m + 1, 1 ≤ #clean1rows(mx, i), ¬ mx[m, 0] < i, ∀ n. n < #c mx → mx[m, n] < i ⊦

Rule argument

Additional information

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

Proof step 44 from proof clean0-plus-clean1-bounded (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < m ↔ m ≠ 0; (s) Used terms: 0