Proof step 1 from proof scolsorted-samecount-above-clean0 (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, #clean0rows(mx0, i), #r mx, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx, λ j. j < i), c < #c mx, scolsorted(mx, #c mx, mx0) ⊦

Rule argument

Additional information

Proof step 2 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, scolsorted(mx, #c mx0, mx0) ⊦

Rule argument

Additional information

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

Proof step 3 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 4:

  ⊦ perm(x, y) → count(x, p) = count(y, p)

• Proof step 5:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, scolsorted(mx, #c mx0, mx0), perm(mx0[c]c, mx[c]c) → count(mx0[c]c, λ j. j < i) = count(mx[c]c, λ j. j < i) ⊦

Rule argument

Additional information

Used terms: mx0[c]c, mx[c]c, λ j. j < i

Proof step 4 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

A spec-lemma

Proof step 5 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 6:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, λ j. j < i) = count(mx[c]c, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, scolsorted(mx, #c mx0, mx0) ⊦

Rule argument

Additional information

Used simplifier rules: scolsorted-permcol: #c mx = #c mx0, c < #c mx ⊦ scolsorted(mx, #c mx0, mx0) → perm(mx0[c]c, mx[c]c);

Proof step 6 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ count(x, p) = count(x, 0, # x, p)

• Proof step 8:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, 0, # mx0[c]c, λ j. j < i) = count(mx[c]c, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, scolsorted(mx, #c mx0, mx0) ⊦

Rule argument

Additional information

Used terms: mx0[c]c, λ j. j < i

Proof step 7 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

A spec-lemma

Proof step 8 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

  ⊦ count(x, p) = count(x, 0, # x, p)

• Proof step 10:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, 0, # mx0[c]c, λ j. j < i) = count(mx[c]c, 0, # mx[c]c, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, scolsorted(mx, #c mx0, mx0) ⊦

Rule argument

Additional information

Used terms: mx[c]c, λ j. j < i

Proof step 9 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

A spec-lemma

Proof step 10 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

  m ≤ n, n ≤ n0, n0 ≤ # x ⊦ count(x, m, n, p) + count(x, n, n0, p) = count(x, m, n0, p)

• Proof step 12:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx0[c]c, #clean0rows(mx0, i), # mx0[c]c, λ j. j < i) = count(mx[c]c, 0, # mx[c]c, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, scolsorted(mx, #c mx0, mx0), 0 ≤ #clean0rows(mx0, i), #clean0rows(mx0, i) ≤ # mx0[c]c, # mx0[c]c ≤ # mx0[c]c ⊦

Rule argument

Additional information

Used simplifier rules: clean0rows-bounded: ⊦ ¬ #r mx < #clean0rows(mx, i); ⊦ 0 ≤ m; (s) ⊦ # mx[c]c = #r mx; (s) ⊦ m ≤ m; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) Used terms: # mx0[c]c, λ j. j < i, #clean0rows(mx0, i), 0, mx0[c]c

Proof step 11 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

A spec-lemma

Proof step 12 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

  m ≤ n, n ≤ # x, ∀ n0. m ≤ n0 ∧ n0 < n → p(x[n0]) ⊦ count(x, m, n, p) = n - m

• Proof step 14:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx0[c]c, #clean0rows(mx0, i), # mx0[c]c, λ j. j < i) = count(mx[c]c, 0, # mx[c]c, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, scolsorted(mx, #c mx0, mx0), # mx0[c]c ≤ # mx0[c]c, #clean0rows(mx0, i) ≤ # mx0[c]c, 0 ≤ #clean0rows(mx0, i) ⊦ 0 ≤ #clean0rows(mx0, i) ∧ #clean0rows(mx0, i) ≤ # mx0[c]c ∧ (∀ n0. 0 ≤ n0 ∧ n0 < #clean0rows(mx0, i) → mx0[c]c[n0] < i)

• Proof step 22:

  #c mx = #c mx0, #r mx = #r mx0, (#clean0rows(mx0, i) - 0) + count(mx0[c]c, #clean0rows(mx0, i), # mx0[c]c, λ j. j < i) = count(mx[c]c, 0, # mx[c]c, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, scolsorted(mx, #c mx0, mx0), # mx0[c]c ≤ # mx0[c]c, 0 ≤ #clean0rows(mx0, i), #clean0rows(mx0, i) ≤ # mx0[c]c, ∀ n0. 0 ≤ n0 ∧ n0 < #clean0rows(mx0, i) → mx0[c]c[n0] < i ⊦

Rule argument

Additional information

Used terms: mx0[c]c, 0, #clean0rows(mx0, i), λ j. j < i

Proof step 13 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

A spec-lemma

Proof step 14 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, n0 < #clean0rows(mx0, i), n0 < #r mx0, scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #r mx0, ¬ mx0[n0, c] < i ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ mx[c]c[r] = mx[r, c]; (s) ⊦ 0 ≤ m; (s) ⊦ # mx[c]c = #r mx; (s) ⊦ m ≤ m; (s)

Proof step 15 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 16:

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

• Proof step 17:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), n0 < #r mx0, n0 < #clean0rows(mx0, i), c < #c mx0, scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #r mx0, ¬ mx0[n0, c] < i, n0 < #clean0rows(mx0, i) → count(mx0[n0]r, λ j. ¬ j < i) = 0 ⊦

Rule argument

Additional information

Used terms: n0, mx0, i

Proof step 16 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Lemma clean0rows-allls

Proof step 17 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #r mx0, λ j. j < i), count(mx0[n0]r, λ j. ¬ j < i) = 0, count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, n0 < #clean0rows(mx0, i), n0 < #r mx0, scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #r mx0, ¬ mx0[n0, c] < i ⊦

Rule argument

Additional information

Proof step 18 from proof scolsorted-samecount-above-clean0 (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:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), n0 < #r mx0, n0 < #clean0rows(mx0, i), c < #c mx0, scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #r mx0, ¬ mx0[n0, c] < i, ∀ n. n < # mx0[n0]r → ¬ ¬ mx0[n0]r[n] < i ⊦

Rule argument

Additional information

Used terms: mx0[n0]r, λ j. ¬ j < i

Proof step 19 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

A spec-lemma

Proof step 20 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  #c mx = #c mx0, #r mx = #r mx0, count(mx0[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, n0 < #clean0rows(mx0, i), n0 < #r mx0, scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #r mx0, ¬ mx0[n0, c] < i, ∀ n. n < #c mx0 → mx0[n0, 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 scolsorted-samecount-above-clean0 (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: c

Proof step 22 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  #c mx = #c mx0, #r mx = #r mx0, #clean0rows(mx0, i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #r mx0, ∀ n0. n0 < #clean0rows(mx0, i) → mx0[n0, c] < i ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ mx[c]c[r] = mx[r, c]; (s) ⊦ 0 ≤ m; (s) ⊦ m ≤ m; (s) ⊦ # mx[c]c = #r mx; (s) ⊦ m - 0 = m; (s)

Proof step 23 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 24:

  m ≤ n, n ≤ n0, n0 ≤ # x ⊦ count(x, m, n, p) + count(x, n, n0, p) = count(x, m, n0, p)

• Proof step 25:

  #c mx = #c mx0, #r mx = #r mx0, #clean0rows(mx0, i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, scolsorted(mx, #c mx0, mx0), 0 ≤ #clean0rows(mx0, i), #clean0rows(mx0, i) ≤ #r mx0, #r mx0 ≤ # mx[c]c, ∀ n0. n0 < #clean0rows(mx0, i) → mx0[n0, c] < i ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s) ⊦ # mx[c]c = #r mx; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m < n → n ≠ 0; (s) ⊦ m < n → ¬ n < m; (s) Used terms: #r mx0, λ j. j < i, #clean0rows(mx0, i), 0, mx[c]c

Proof step 24 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

A spec-lemma

Proof step 25 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  m ≤ n, n ≤ # x, ∀ n0. m ≤ n0 ∧ n0 < n → p(x[n0]) ⊦ count(x, m, n, p) = n - m

• Proof step 27:

  #c mx = #c mx0, #r mx = #r mx0, #clean0rows(mx0, i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, scolsorted(mx, #c mx0, mx0), #r mx0 ≤ # mx[c]c, #clean0rows(mx0, i) ≤ #r mx0, 0 ≤ #clean0rows(mx0, i), ∀ n0. n0 < #clean0rows(mx0, i) → mx0[n0, c] < i ⊦ 0 ≤ #clean0rows(mx0, i) ∧ #clean0rows(mx0, i) ≤ # mx[c]c ∧ (∀ n0. 0 ≤ n0 ∧ n0 < #clean0rows(mx0, i) → mx[c]c[n0] < i)

Rule argument

Additional information

Used simplifier rules: ⊦ m - 0 = m; (s) ⊦ m + n = m + n0 ↔ n = n0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: mx[c]c, 0, #clean0rows(mx0, i), λ j. j < i

Proof step 26 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

A spec-lemma

Proof step 27 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  #c mx = #c mx0, #r mx = #r mx0, #clean0rows(mx0, i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, n0 < #clean0rows(mx0, i), n0 < #r mx0, scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #r mx0, ¬ mx[n0, c] < i, ∀ n0. n0 < #clean0rows(mx0, i) → mx0[n0, c] < i ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m ≤ m; (s) let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ mx[c]c[r] = mx[r, c]; (s) ⊦ 0 ≤ m; (s) ⊦ # mx[c]c = #r mx; (s)

Proof step 28 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 29:

  #c mx ≠ 0, #r mx = #r mx0, #c mx = #c mx0 ⊦ scolsorted(mx, #c mx0, mx0) → ¬ #clean0rows(mx, i) < #clean0rows(mx0, i)

• Proof step 30:

  #c mx = #c mx0, #r mx = #r mx0, #clean0rows(mx0, i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), n0 < #r mx0, n0 < #clean0rows(mx0, i), c < #c mx0, scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #r mx0, ¬ mx[n0, c] < i, scolsorted(mx, #c mx0, mx0) → ¬ #clean0rows(mx, i) < #clean0rows(mx0, i), ∀ n0. n0 < #clean0rows(mx0, i) → mx0[n0, c] < i ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic Used terms: mx, mx0, i

Proof step 29 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Lemma scolsorted-clean0notls

Proof step 30 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  #c mx = #c mx0, #r mx = #r mx0, #clean0rows(mx0, i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, n0 < #clean0rows(mx0, i), n0 < #r mx0, n0 < #clean0rows(mx, i), scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #r mx0, #clean0rows(mx0, i) ≤ #clean0rows(mx, i), ¬ mx[n0, c] < i, ∀ n0. n0 < #clean0rows(mx0, i) → mx0[n0, c] < i ⊦

Rule argument

Additional information

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

Proof step 31 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 32:

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

• Proof step 33:

  #c mx = #c mx0, #r mx = #r mx0, #clean0rows(mx0, i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), n0 < #clean0rows(mx, i), n0 < #r mx0, n0 < #clean0rows(mx0, i), c < #c mx0, scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #clean0rows(mx, i), #clean0rows(mx0, i) ≤ #r mx0, ¬ mx[n0, c] < i, n0 < #clean0rows(mx, i) → count(mx[n0]r, λ j. ¬ j < i) = 0, ∀ n0. n0 < #clean0rows(mx0, i) → mx0[n0, c] < i ⊦

Rule argument

Additional information

Used terms: n0, mx, i

Proof step 32 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Lemma clean0rows-allls

Proof step 33 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  #c mx = #c mx0, #r mx = #r mx0, #clean0rows(mx0, i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), count(mx[n0]r, λ j. ¬ j < i) = 0, count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, n0 < #clean0rows(mx0, i), n0 < #r mx0, n0 < #clean0rows(mx, i), scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #r mx0, #clean0rows(mx0, i) ≤ #clean0rows(mx, i), ¬ mx[n0, c] < i, ∀ n0. n0 < #clean0rows(mx0, i) → mx0[n0, c] < i ⊦

Rule argument

Additional information

Proof step 34 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 35:

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

• Proof step 36:

  #c mx = #c mx0, #r mx = #r mx0, #clean0rows(mx0, i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), n0 < #clean0rows(mx, i), n0 < #r mx0, n0 < #clean0rows(mx0, i), c < #c mx0, scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #clean0rows(mx, i), #clean0rows(mx0, i) ≤ #r mx0, ¬ mx[n0, c] < i, ∀ n0. n0 < #clean0rows(mx0, i) → mx0[n0, c] < i, ∀ n. n < # mx[n0]r → ¬ ¬ mx[n0]r[n] < i ⊦

Rule argument

Additional information

Used terms: mx[n0]r, λ j. ¬ j < i

Proof step 35 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

A spec-lemma

Proof step 36 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  #c mx = #c mx0, #r mx = #r mx0, #clean0rows(mx0, i) + count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) = count(mx[c]c, 0, #clean0rows(mx0, i), λ j. j < i) + count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), count(mx0[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i) ≠ count(mx[c]c, #clean0rows(mx0, i), #r mx0, λ j. j < i), c < #c mx0, n0 < #clean0rows(mx0, i), n0 < #r mx0, n0 < #clean0rows(mx, i), scolsorted(mx, #c mx0, mx0), #clean0rows(mx0, i) ≤ #r mx0, #clean0rows(mx0, i) ≤ #clean0rows(mx, i), ¬ mx[n0, c] < i, ∀ n0. n0 < #clean0rows(mx0, i) → mx0[n0, c] < i, ∀ n. n < #c mx0 → mx[n0, 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 37 from proof scolsorted-samecount-above-clean0 (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: c