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Proof step 1 from proof dirtyrowsone-rowsincreasing (conclusion) (back to tree)

#r mx = #r mx0, #c mx = #c mx0 ⊦ mx.dirtyrows≤(1) ∧ scolsorted(mx, #c mx, mx0) → rowsincreasing(mx, #r mx)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, mx.dirtyrows≤(1), scolsorted(mx, #c mx, mx0), ¬ rowsincreasing(mx, #r mx) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 3 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, mx.dirtyrows≤(1), scolsorted(mx, #c mx0, mx0), ¬ rowsincreasing(mx, #r mx0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:dirtyrowsbound-def, seq: ⊦ mx.dirtyrows≤(bound) ↔ (∀ i. #dirtyrows(mx, i) ≤ bound), subst:[mx, bound] → [mx, 1], <all paths>

Additional information

Used terms: mx, 1

Proof step 4 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

⊦ mx.dirtyrows≤(bound) ↔ (∀ i. #dirtyrows(mx, i) ≤ bound)

Lemma dirtyrowsbound-def

Proof step 5 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, scolsorted(mx, #c mx0, mx0), ¬ rowsincreasing(mx, #r mx0), ∀ i. #dirtyrows(mx, i) ≤ 1 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:rowsincreasing-def, seq:⊦ rowsincreasing(mx, r) ↔ (∀ r1, r2, c1, c2. r1 < r2 ∧ r2 < r ∧ c1 < #c mx ∧ c2 < #c mx → ¬ mx[r2, c2] < mx[r1, c1]), subst:[mx, r] → [mx, #r mx0], <all paths>

Additional information

Used terms: mx, #r mx0

Proof step 6 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

⊦ rowsincreasing(mx, r) ↔ (∀ r1, r2, c1, c2. r1 < r2 ∧ r2 < r ∧ c1 < #c mx ∧ c2 < #c mx → ¬ mx[r2, c2] < mx[r1, c1])

Lemma rowsincreasing-def

Proof step 7 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, scolsorted(mx, #c mx0, mx0), ∀ i. #dirtyrows(mx, i) ≤ 1, ∃ r1, r2, c1, c2. ¬(r1 < r2 ∧ r2 < #r mx0 ∧ c1 < #c mx ∧ c2 < #c mx → ¬ mx[r2, c2] < mx[r1, c1]) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic ⊦ m +1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) trans-nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 8 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, scolsorted(mx, #c mx0, mx0), ∀ i. #dirtyrows(mx, i) ≤ 1 ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:dirytrows-leone, seq:#c mx ≠ 0, #dirtyrows(mx, i) ≤ 1 ⊦ (∀ r. r < #r mx → count(mx[r]r, λ j. j < i) = 0 ∨ count(mx[r]r, λ j. ¬ j < i) = 0) ∨ (∃ r0. count(mx[r0]r, λ j. j < i) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < i) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx → count(mx[r]r, λ j. j < i) = 0 ∨ count(mx[r]r, λ j. ¬ j < i) = 0)), subst:[mx, i] → [mx, mx[r1, c1]]

Additional information

Used terms: mx, mx[r1, c1]

Proof step 9 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx ≠ 0, #dirtyrows(mx, i) ≤ 1 ⊦ (∀ r. r < #r mx → count(mx[r]r, λ j. j < i) = 0 ∨ count(mx[r]r, λ j. ¬ j < i) = 0) ∨ (∃ r0. count(mx[r0]r, λ j. j < i) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < i) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx → count(mx[r]r, λ j. j < i) = 0 ∨ count(mx[r]r, λ j. ¬ j < i) = 0))

Lemma dirytrows-leone

Proof step 10 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, scolsorted(mx, #c mx0, mx0), ∀ i. #dirtyrows(mx, i) ≤ 1 ⊦ #c mx ≠ 0 ∧ #dirtyrows(mx, mx[r1, c1]) ≤ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < n → n ≠ 0; (s)

Proof step 11 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, #dirtyrows(mx, mx[r1, c1]) ≠ 0, #dirtyrows(mx, mx[r1, c1]) ≠ 1, r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, scolsorted(mx, #c mx0, mx0), ∀ i. #dirtyrows(mx, i) ≤ 1 ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 13 terms: mx[r1, c1] abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); (s) Used terms: mx[r1, c1]

Proof step 12 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, scolsorted(mx, #c mx0, mx0), ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ r. r < #r mx → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ (∃ r0. count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 13 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, scolsorted(mx, #c mx0, mx0), ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:scolsorted-def, seq:⊦ m ≤ #c mx → ( scolsorted(mx, m, mx0) ↔ #r mx = #r mx0 ∧ #c mx = #c mx0 ∧ (∀ c. c < #c mx → (c < m ⊃ perm(mx0[c]c, mx[c]c) ∧ ordered≤(mx[c]c) ;mx0[c]c = mx[c]c ))), subst:[mx, m, mx0] → [mx, #c mx0, mx0], <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss) 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 lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic trans-nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic ⊦ m < n → ¬ n < m; (s) Used terms: mx, #c mx0, mx0

Proof step 14 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

⊦ m ≤ #c mx → ( scolsorted(mx, m, mx0) ↔ #r mx = #r mx0 ∧ #c mx = #c mx0 ∧ (∀ c. c < #c mx → (c < m ⊃ perm(mx0[c]c, mx[c]c) ∧ ordered≤(mx[c]c);mx0[c]c = mx[c]c)))

Lemma scolsorted-def

Proof step 15 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, #c mx0 ≤ #c mx, ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ c. c < #c mx → (c < #c mx0 ⊃ perm(mx0[c]c, mx[c]c) ∧ ordered≤(mx[c]c);mx0[c]c = mx[c]c), (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat

Proof step 16 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ c. c < #c mx0 → perm(mx0[c]c, mx[c]c) ∧ ordered≤(mx[c]c), (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 11 terms: c1 abort: F discard: F computed: T

Additional information

Used terms: c1

Proof step 17 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, c1 < #c mx0 → perm(mx0[c1]c, mx[c1]c) ∧ ordered≤(mx[c1]c), ∀ c. c < #c mx0 → perm(mx0[c]c, mx[c]c) ∧ ordered≤(mx[c]c), ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 18 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, ordered≤(mx[c1]c), perm(mx0[c1]c, mx[c1]c), ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ c. c < #c mx0 → perm(mx0[c]c, mx[c]c) ∧ ordered≤(mx[c]c), (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 13 terms: c2 abort: F discard: T computed: T

Additional information

Used terms: c2

Proof step 19 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, ordered≤(mx[c1]c), perm(mx0[c1]c, mx[c1]c), c2 < #c mx0 → perm(mx0[c2]c, mx[c2]c) ∧ ordered≤(mx[c2]c), ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 20 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, ordered≤(mx[c1]c), ordered≤(mx[c2]c), perm(mx0[c1]c, mx[c1]c), perm(mx0[c2]c, mx[c2]c), ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

User applied rule weakening

And obtained the following new goals

Rule argument

poslist: left 12, left 13

Additional information

Proof step 21 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, ordered≤(mx[c2]c), ordered≤(mx[c1]c), ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: olist-new[oarray-sort-new], name:orderedle-all, seq: ⊦ ordered≤(x) ↔ (∀ m, n. m ≤ n ∧ n < # x → ¬ x[n] < x[m]), subst:[x] → [mx[c1]c], <all paths>

Additional information

Used terms: mx[c1]c

Proof step 22 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

⊦ ordered≤(x) ↔ (∀ m, n. m ≤ n ∧ n < # x → ¬ x[n] < x[m])

A spec-lemma

Proof step 23 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, ordered≤(mx[c2]c), ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ m, n. m ≤ n ∧ n < # mx[c1]c → ¬ mx[c1]c[n] < mx[c1]c[m], (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 24 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, ordered≤(mx[c2]c), ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ m, n. m ≤ n ∧ n < #r mx0 → ¬ mx[n, c1] < mx[m, c1], (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 12 terms: r1, r2 abort: F discard: T computed: F

Additional information

Used terms: r1, r2

Proof step 25 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, ordered≤(mx[c2]c), r1 ≤ r2 ∧ r2 < #r mx0 → ¬ mx[r2, c1] < mx[r1, c1], ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 26 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, ordered≤(mx[c2]c), ¬ mx[r2, c1] < mx[r1, c1], ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: olist-new[oarray-sort-new], name:orderedle-all, seq: ⊦ ordered≤(x) ↔ (∀ m, n. m ≤ n ∧ n < # x → ¬ x[n] < x[m]), subst:[x] → [mx[c2]c], <all paths>

Additional information

Used terms: mx[c2]c

Proof step 27 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

⊦ ordered≤(x) ↔ (∀ m, n. m ≤ n ∧ n < # x → ¬ x[n] < x[m])

A spec-lemma

Proof step 28 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, ¬ mx[r2, c1] < mx[r1, c1], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ m, n. m ≤ n ∧ n < # mx[c2]c → ¬ mx[c2]c[n] < mx[c2]c[m], (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 29 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, ¬ mx[r2, c1] < mx[r1, c1], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ m, n. m ≤ n ∧ n < #r mx0 → ¬ mx[n, c2] < mx[m, c2], (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 12 terms: r1, r2 abort: F discard: T computed: F

Additional information

Used terms: r1, r2

Proof step 30 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, ¬ mx[r2, c1] < mx[r1, c1], r1 ≤ r2 ∧ r2 < #r mx0 → ¬ mx[r2, c2] < mx[r1, c2], ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem ⊦ m < n → m ≤ n; (s)

Proof step 31 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 13 terms: mx[r1, c1] abort: F discard: F computed: F

Additional information

Used terms: mx[r1, c1]

Proof step 32 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ∨ count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0 ∧ count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0 ∧ (∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0) ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 15

Additional information

Proof step 33 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0 ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 15 terms: r2 abort: F discard: F computed: F

Additional information

Used terms: r2

Proof step 34 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], r2 < #r mx0 → count(mx[r2]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 35 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, count(mx[r2]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: filter-count[oarray-sort-new], name:count-zero, seq: ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n])), subst:[x, p] → [mx[r2]r, λ j. j < mx[r1, c1]], <all paths>

Additional information

Used terms: mx[r2]r, λ j. j < mx[r1, c1]

Proof step 36 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

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

A spec-lemma

Proof step 37 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ n. n < # mx[r2]r → ¬ mx[r2]r[n] < mx[r1, c1]) ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 38 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, (∀ n. n < #c mx0 → ¬ mx[r2, n] < mx[r1, c1]) ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: filter-count[oarray-sort-new], name:count-zero, seq: ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n])), subst:[x, p] → [mx[r2]r, λ j. ¬ j < mx[r1, c1]], <all paths>

Additional information

Used terms: mx[r2]r, λ j. ¬ j < mx[r1, c1]

Proof step 39 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

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

A spec-lemma

Proof step 40 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ n. n < #c mx0 → ¬ mx[r2, n] < mx[r1, c1]) ∨ (∀ n. n < # mx[r2]r → ¬ ¬ mx[r2]r[n] < mx[r1, c1]) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 41 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, (∀ n. n < #c mx0 → ¬ mx[r2, n] < mx[r1, c1]) ∨ (∀ n. n < #c mx0 → mx[r2, n] < mx[r1, c1]) ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 16

Additional information

Proof step 42 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ n. n < #c mx0 → ¬ mx[r2, n] < mx[r1, c1] ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 16 terms: c2

Additional information

Used terms: c2

Proof step 43 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ∀ r. r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ n. n < #c mx0 → mx[r2, n] < mx[r1, c1] ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 16 terms: c1

Additional information

Used terms: c1

Proof step 44 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0, count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0 ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 17 terms: r2 abort: F discard: F computed: T

Additional information

Used terms: r2

Proof step 45 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0, count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], r2 ≠ r0 ∧ r2 < #r mx0 → count(mx[r2]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 46 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, count(mx[r0]r, λ j. ¬ j < mx[r1, c1]) ≠ 0, count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], r2 ≠ r0 → count(mx[r2]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: filter-count[oarray-sort-new], name:count-zero, seq: ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n])), subst:[x, p] → [mx[r0]r, λ j. ¬ j < mx[r1, c1]], <all paths>

Additional information

Used terms: mx[r0]r, λ j. ¬ j < mx[r1, c1]

Proof step 47 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

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

A spec-lemma

Proof step 48 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], r2 ≠ r0 → count(mx[r2]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, ∃ n. ¬(n < # mx[r0]r → ¬ ¬ mx[r0]r[n] < mx[r1, c1]) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 49 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, count(mx[r0]r, λ j. j < mx[r1, c1]) ≠ 0, n < #c mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r0, n] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ¬ mx[r2, c1] < mx[r1, c1], r2 ≠ r0 → count(mx[r2]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: filter-count[oarray-sort-new], name:count-zero, seq: ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n])), subst:[x, p] → [mx[r0]r, λ j. j < mx[r1, c1]], <all paths>

Additional information

Used terms: mx[r0]r, λ j. j < mx[r1, c1]

Proof step 50 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

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

A spec-lemma

Proof step 51 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], n < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r0, n] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ¬ mx[r2, c1] < mx[r1, c1], r2 ≠ r0 → count(mx[r2]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∃ n. ¬(n < # mx[r0]r → ¬ mx[r0]r[n] < mx[r1, c1]) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 52 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, n0 < #c mx0, mx[r0, n0] < mx[r1, c1], r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], n < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], r2 ≠ r0 → count(mx[r2]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0 ⊦

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: r2 ≠ r0

Additional information

Used formulas: r2 ≠ r0

Proof step 53 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r2 ≠ r0, n < #c mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, n0] < mx[r1, c1], n0 < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], r2 ≠ r0 → count(mx[r2]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 54 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r2 ≠ r0, n0 < #c mx0, mx[r0, n0] < mx[r1, c1], r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], n < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, count(mx[r2]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0 ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 21

Additional information

Proof step 55 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, count(mx[r2]r, λ j. j < mx[r1, c1]) = 0, r2 ≠ r0, n < #c mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, n0] < mx[r1, c1], n0 < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: filter-count[oarray-sort-new], name:count-zero, seq: ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n])), subst:[x, p] → [mx[r2]r, λ j. j < mx[r1, c1]], <all paths>

Additional information

Used terms: mx[r2]r, λ j. j < mx[r1, c1]

Proof step 56 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

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

A spec-lemma

Proof step 57 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r2 ≠ r0, n0 < #c mx0, mx[r0, n0] < mx[r1, c1], r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], n < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ n. n < # mx[r2]r → ¬ mx[r2]r[n] < mx[r1, c1] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 58 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r2 ≠ r0, n < #c mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, n0] < mx[r1, c1], n0 < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ n1. n1 < #c mx0 → ¬ mx[r2, n1] < mx[r1, c1] ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 21 terms: c2

Additional information

Used terms: c2

Proof step 59 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0, r2 ≠ r0, n < #c mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, n0] < mx[r1, c1], n0 < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: filter-count[oarray-sort-new], name:count-zero, seq: ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n])), subst:[x, p] → [mx[r2]r, λ j. ¬ j < mx[r1, c1]], <all paths>

Additional information

Used terms: mx[r2]r, λ j. ¬ j < mx[r1, c1]

Proof step 60 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

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

A spec-lemma

Proof step 61 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r2 ≠ r0, n0 < #c mx0, mx[r0, n0] < mx[r1, c1], r1 + 1 < #r mx0, mx[r2, c2] < mx[r1, c1], r1 < r2, r2 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], n < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ n. n < # mx[r2]r → ¬ ¬ mx[r2]r[n] < mx[r1, c1] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 62 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r2 ≠ r0, n < #c mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, n0] < mx[r1, c1], n0 < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ n1. n1 < #c mx0 → mx[r2, n1] < mx[r1, c1] ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 21 terms: c1

Additional information

Used terms: c1

Proof step 63 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r2 = r0, n < #c mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r2 < #r mx0, r1 < r2, mx[r2, c2] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, n0] < mx[r1, c1], n0 < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r2, c1] < mx[r1, c1], ¬ mx[r2, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], r2 ≠ r0 → count(mx[r2]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r2]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 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 lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic

Proof step 64 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, n < #c mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r0 < #r mx0, r1 < r0, mx[r0, c2] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, n0] < mx[r1, c1], n0 < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r0, n] < mx[r1, c1], ¬ mx[r0, c2] < mx[r1, c2], ¬ mx[r0, c1] < mx[r1, c1], ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1 ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 18 terms: r1 abort: F discard: F computed: T

Additional information

Used terms: r1

Proof step 65 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, n0 < #c mx0, mx[r0, n0] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, c2] < mx[r1, c1], r1 < r0, r0 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], n < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r0, n] < mx[r1, c1], ¬ mx[r0, c2] < mx[r1, c2], ¬ mx[r0, c1] < mx[r1, c1], r1 ≠ r0 ∧ r1 < #r mx0 → count(mx[r1]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r1]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)

Proof step 66 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 ≠ r0, n0 < #c mx0, mx[r0, n0] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, c2] < mx[r1, c1], r1 < r0, r0 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], n < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r0, c1] < mx[r1, c1], ¬ mx[r0, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, count(mx[r1]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r1]r, λ j. ¬ j < mx[r1, c1]) = 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: filter-count[oarray-sort-new], name:count-zero, seq: ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n])), subst:[x, p] → [mx[r1]r, λ j. j < mx[r1, c1]], <all paths>

Additional information

Used terms: mx[r1]r, λ j. j < mx[r1, c1]

Proof step 67 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

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

A spec-lemma

Proof step 68 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 ≠ r0, n < #c mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r0 < #r mx0, r1 < r0, mx[r0, c2] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, n0] < mx[r1, c1], n0 < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r0, c1] < mx[r1, c1], ¬ mx[r0, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, (∀ n. n < # mx[r1]r → ¬ mx[r1]r[n] < mx[r1, c1]) ∨ count(mx[r1]r, λ j. ¬ j < mx[r1, c1]) = 0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: filter-count[oarray-sort-new], name:count-zero, seq: ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n])), subst:[x, p] → [mx[r1]r, λ j. ¬ j < mx[r1, c1]], <all paths>

Additional information

Used terms: mx[r1]r, λ j. ¬ j < mx[r1, c1]

Proof step 69 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

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

A spec-lemma

Proof step 70 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 ≠ r0, n0 < #c mx0, mx[r0, n0] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, c2] < mx[r1, c1], r1 < r0, r0 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], n < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r0, c1] < mx[r1, c1], ¬ mx[r0, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, (∀ n. n < # mx[r1]r → ¬ mx[r1]r[n] < mx[r1, c1]) ∨ (∀ n. n < # mx[r1]r → ¬ ¬ mx[r1]r[n] < mx[r1, c1]) ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 21

Additional information

Proof step 71 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 ≠ r0, n < #c mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r0 < #r mx0, r1 < r0, mx[r0, c2] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, n0] < mx[r1, c1], n0 < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r0, c1] < mx[r1, c1], ¬ mx[r0, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ n. n < # mx[r1]r → ¬ mx[r1]r[n] < mx[r1, c1] ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 21 terms: c2 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ # mx[r]r = #c mx; (s) ⊦ mx[r]r[c] = mx[r, c]; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ ¬ n < n; (s) Used terms: c2

Proof step 72 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 ≠ r0, n < #c mx0, mx[r1, c2] < mx[r1, c1], r1 < #r mx0, c2 < #c mx0, c1 < #c mx0, r0 < #r mx0, r1 < r0, mx[r0, c2] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, n0] < mx[r1, c1], n0 < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r0, c1] < mx[r1, c1], ¬ mx[r0, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ n. n < # mx[r1]r → ¬ ¬ mx[r1]r[n] < mx[r1, c1] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 73 from proof dirtyrowsone-rowsincreasing (previous step) (back to tree)

#c mx = #c mx0, #r mx = #r mx0, r1 ≠ r0, n0 < #c mx0, mx[r0, n0] < mx[r1, c1], r1 + 1 < #r mx0, mx[r0, c2] < mx[r1, c1], r1 < r0, r0 < #r mx0, c1 < #c mx0, c2 < #c mx0, r1 < #r mx0, mx[r1, c2] < mx[r1, c1], n < #c mx0, #dirtyrows(mx, mx[r1, c1]) ≤ 1, ¬ mx[r0, c1] < mx[r1, c1], ¬ mx[r0, c2] < mx[r1, c2], ¬ mx[r0, n] < mx[r1, c1], ∀ i. #dirtyrows(mx, i) ≤ 1, ∀ r. r ≠ r0 ∧ r < #r mx0 → count(mx[r]r, λ j. j < mx[r1, c1]) = 0 ∨ count(mx[r]r, λ j. ¬ j < mx[r1, c1]) = 0, ∀ n1. n1 < #c mx0 → mx[r1, n1] < mx[r1, c1] ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 21 terms: c1

Additional information

Used simplifier rules: ⊦ ¬ a < a; (s) Used terms: c1

Proof Tree for dirtyrowsone-rowsincreasing (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