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Proof step 1 from proof count1-leq-clean1-ind (conclusion) (back to tree)

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

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof count1-leq-clean1-ind (previous step) (back to tree)

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: nat, name:elim-sub, seq:n ≤ m ⊦ n0 = m - n ↔ m = n0 + n, subst:[m, n, n0] → [#r mx, #clean1rows(mx, i), n0]

Additional information

Used simplifier rules: clean1rows-bounded: ⊦ ¬ #r mx < #clean1rows(mx, i); ⊦ m ≤ n ↔ ¬ n < m; (ss) n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: #r mx, #clean1rows(mx, i), n0

Proof step 3 from proof count1-leq-clean1-ind (previous step) (back to tree)

#r mx = n0 + #clean1rows(mx, i), count(mx[c]c, n0, n0 + m, λ j. ¬ j < i) ≠ m, c < #c mx, m ≤ #clean1rows(mx, i), #clean1rows(mx, i) ≤ #r mx ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 4 from proof count1-leq-clean1-ind (previous step) (back to tree)

#r mx = n0 + #clean1rows(mx, i), count(mx[c]c, n0, n0 + m, λ j. ¬ j < i) ≠ m, c < #c mx, m ≤ #r mx, m ≤ #clean1rows(mx, i), #clean1rows(mx, i) ≤ #r mx ⊦

User applied rule structural induction

And obtained the following new goals

Rule argument

term: m

Additional information

Used simplifier rules: nat freely generated by 0, +1;; ⊦ n + 0 = n; (s) ⊦ # mx[c]c = #r mx; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ 0 ≤ m; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic m ≤ # x ⊦ count(x, m, m, p) = 0; (s)

Proof step 5 from proof count1-leq-clean1-ind (previous step) (back to tree)

#r mx = n0 + #clean1rows(mx, i), count(mx[c]c, n0, n0 + m0 +1, λ j. ¬ j < i) ≠ m0 +1, c < #c mx, #clean1rows(mx, i) ≤ #r mx, m0 +1 ≤ #clean1rows(mx, i), m0 +1 ≤ #r mx, ∀ mx, n0, i, c. ¬ ( #r mx = n0 + #clean1rows(mx, i) ∧ count(mx[c]c, n0, n0 + m0, λ j. ¬ j < i) ≠ m0 ∧ c < #c mx ∧ m0 ≤ #r mx ∧ m0 ≤ #clean1rows(mx, i) ∧ #clean1rows(mx, i) ≤ #r mx) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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 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 +1 = m + 1; (s) ⊦ mx[c]c[r] = mx[r, c]; (s) m ≤ n, n ≤ # x ⊦ count(x, m, n + 1, p) = count(x, m, n, p) + (p(x[n]) ⊃ 1;0); (s) ⊦ m = m0 + n0 ∧ n < n0 → ¬ m < m0 + n; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ # mx[c]c = #r mx; (s) ⊦ m ≤ m + n; (s)

Proof step 6 from proof count1-leq-clean1-ind (previous step) (back to tree)

#r mx = n0 + #clean1rows(mx, i), count(mx[c]c, n0, n0 + m0, λ j. ¬ j < i) + (¬ mx[n0 + m0, c] < i ⊃ 1;0) ≠ m0 + 1, m0 < #clean1rows(mx, i), m0 < #r mx, c < #c mx, #clean1rows(mx, i) ≤ #r mx, ∀ mx0, n, i0, c0. #r mx0 ≠ n + #clean1rows(mx0, i0) ∨ count(mx0[c0]c, n, n + m0, λ j. ¬ j < i0) = m0 ∨ ¬ c0 < #c mx0 ∨ ¬ m0 ≤ #r mx0 ∨ ¬ m0 ≤ #clean1rows(mx0, i0) ∨ ¬ #clean1rows(mx0, i0) ≤ #r mx0 ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 2

Additional information

Proof step 7 from proof count1-leq-clean1-ind (previous step) (back to tree)

#r mx = n0 + #clean1rows(mx, i), count(mx[c]c, n0, n0 + m0, λ j. ¬ j < i) + 1 ≠ m0 + 1, c < #c mx, m0 < #r mx, m0 < #clean1rows(mx, i), #clean1rows(mx, i) ≤ #r mx, ¬ mx[n0 + m0, c] < i, ∀ mx0, n, i0, c0. #r mx0 ≠ n + #clean1rows(mx0, i0) ∨ count(mx0[c0]c, n, n + m0, λ j. ¬ j < i0) = m0 ∨ ¬ c0 < #c mx0 ∨ ¬ m0 ≤ #r mx0 ∨ ¬ m0 ≤ #clean1rows(mx0, i0) ∨ ¬ #clean1rows(mx0, i0) ≤ #r mx0 ⊦

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 + n0 ↔ n = n0; (s)

Proof step 8 from proof count1-leq-clean1-ind (previous step) (back to tree)

#r mx = n0 + #clean1rows(mx, i), count(mx[c]c, n0, n0 + m0, λ j. ¬ j < i) ≠ m0, m0 < #clean1rows(mx, i), m0 < #r mx, c < #c mx, #clean1rows(mx, i) ≤ #r mx, ¬ mx[n0 + m0, c] < i, ∀ mx0, n, i0, c0. #r mx0 ≠ n + #clean1rows(mx0, i0) ∨ count(mx0[c0]c, n, n + m0, λ j. ¬ j < i0) = m0 ∨ ¬ c0 < #c mx0 ∨ ¬ m0 ≤ #r mx0 ∨ ¬ m0 ≤ #clean1rows(mx0, i0) ∨ ¬ #clean1rows(mx0, i0) ≤ #r mx0 ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: mx, n0, i, c abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) Used terms: mx, n0, i, c

Proof step 9 from proof count1-leq-clean1-ind (previous step) (back to tree)

#r mx = n0 + #clean1rows(mx, i), count(mx[c]c, n0, n0 + m0, λ j. ¬ j < i) + 0 ≠ m0 + 1, c < #c mx, m0 < #r mx, m0 < #clean1rows(mx, i), mx[n0 + m0, c] < i, #clean1rows(mx, i) ≤ #r mx, ∀ mx0, n, i0, c0. #r mx0 ≠ n + #clean1rows(mx0, i0) ∨ count(mx0[c0]c, n, n + m0, λ j. ¬ j < i0) = m0 ∨ ¬ c0 < #c mx0 ∨ ¬ m0 ≤ #r mx0 ∨ ¬ m0 ≤ #clean1rows(mx0, i0) ∨ ¬ #clean1rows(mx0, i0) ≤ #r mx0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n + 0 = n; (s)

Proof step 10 from proof count1-leq-clean1-ind (previous step) (back to tree)

#r mx = n0 + #clean1rows(mx, i), count(mx[c]c, n0, n0 + m0, λ j. ¬ j < i) ≠ m0 + 1, mx[n0 + m0, c] < i, m0 < #clean1rows(mx, i), m0 < #r mx, c < #c mx, #clean1rows(mx, i) ≤ #r mx, ∀ mx0, n, i0, c0. #r mx0 ≠ n + #clean1rows(mx0, i0) ∨ count(mx0[c0]c, n, n + m0, λ j. ¬ j < i0) = m0 ∨ ¬ c0 < #c mx0 ∨ ¬ m0 ≤ #r mx0 ∨ ¬ m0 ≤ #clean1rows(mx0, i0) ∨ ¬ #clean1rows(mx0, i0) ≤ #r mx0 ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:clean1rows-allge, seq:⊦ #r mx ≤ #clean1rows(mx, i) + r ∧ r < #r mx → count(mx[r]r, λ j. j < i) = 0, subst:[mx, i, r] → [mx, i, n0 + m0]

Additional information

Used terms: mx, i, n0 + m0

Proof step 11 from proof count1-leq-clean1-ind (previous step) (back to tree)

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

Lemma clean1rows-allge

Proof step 12 from proof count1-leq-clean1-ind (previous step) (back to tree)

#r mx = n0 + #clean1rows(mx, i), count(mx[c]c, n0, n0 + m0, λ j. ¬ j < i) ≠ m0 + 1, c < #c mx, m0 < #r mx, m0 < #clean1rows(mx, i), mx[n0 + m0, c] < i, #clean1rows(mx, i) ≤ #r mx, #r mx ≤ #clean1rows(mx, i) + n0 + m0 ∧ n0 + m0 < #r mx → count(mx[n0 + m0]r, λ j. j < i) = 0, ∀ mx0, n, i0, c0. #r mx0 ≠ n + #clean1rows(mx0, i0) ∨ count(mx0[c0]c, n, n + m0, λ j. ¬ j < i0) = m0 ∨ ¬ c0 < #c mx0 ∨ ¬ m0 ≤ #r mx0 ∨ ¬ m0 ≤ #clean1rows(mx0, i0) ∨ ¬ #clean1rows(mx0, i0) ≤ #r mx0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) 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 + n; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ ¬ m + n < m; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s)

Proof step 13 from proof count1-leq-clean1-ind (previous step) (back to tree)

#r mx = n0 + #clean1rows(mx, i), count(mx[n0 + m0]r, λ j. j < i) = 0, count(mx[c]c, n0, n0 + m0, λ j. ¬ j < i) ≠ m0 + 1, c < #c mx, m0 < #r mx, m0 < #clean1rows(mx, i), mx[n0 + m0, c] < i, #clean1rows(mx, i) ≤ #r mx, #r mx ≤ #clean1rows(mx, i) + n0 + m0, ∀ mx0, n, i0, c0. #r mx0 ≠ n + #clean1rows(mx0, i0) ∨ count(mx0[c0]c, n, n + m0, λ j. ¬ j < i0) = m0 ∨ ¬ c0 < #c mx0 ∨ ¬ m0 ≤ #r mx0 ∨ ¬ m0 ≤ #clean1rows(mx0, i0) ∨ ¬ #clean1rows(mx0, i0) ≤ #r mx0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: filter-count[oarray-sort], name:count-zero, seq: ⊦ count(x, p) = 0 ↔ (∀ n. n < # x → ¬ p(x[n])), subst:[x, p] → [mx[n0 + m0]r, λ j. j < i]; ['a] → [tord], <all paths>

Additional information

Used terms: mx[n0 + m0]r, λ j. j < i

Proof step 14 from proof count1-leq-clean1-ind (previous step) (back to tree)

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

A spec-lemma

Proof step 15 from proof count1-leq-clean1-ind (previous step) (back to tree)

#r mx = n0 + #clean1rows(mx, i), count(mx[c]c, n0, n0 + m0, λ j. ¬ j < i) ≠ m0 + 1, mx[n0 + m0, c] < i, m0 < #clean1rows(mx, i), m0 < #r mx, c < #c mx, #r mx ≤ #clean1rows(mx, i) + n0 + m0, #clean1rows(mx, i) ≤ #r mx, ∀ mx0, n, i0, c0. #r mx0 ≠ n + #clean1rows(mx0, i0) ∨ count(mx0[c0]c, n, n + m0, λ j. ¬ j < i0) = m0 ∨ ¬ c0 < #c mx0 ∨ ¬ m0 ≤ #r mx0 ∨ ¬ m0 ≤ #clean1rows(mx0, i0) ∨ ¬ #clean1rows(mx0, i0) ≤ #r mx0, ∀ n. n < # mx[n0 + m0]r → ¬ mx[n0 + m0]r[n] < i ⊦

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 16 from proof count1-leq-clean1-ind (previous step) (back to tree)

#r mx = n0 + #clean1rows(mx, i), count(mx[c]c, n0, n0 + m0, λ j. ¬ j < i) ≠ m0 + 1, c < #c mx, m0 < #r mx, m0 < #clean1rows(mx, i), mx[n0 + m0, c] < i, #clean1rows(mx, i) ≤ #r mx, #r mx ≤ #clean1rows(mx, i) + n0 + m0, ∀ mx0, n, i0, c0. #r mx0 ≠ n + #clean1rows(mx0, i0) ∨ count(mx0[c0]c, n, n + m0, λ j. ¬ j < i0) = m0 ∨ ¬ c0 < #c mx0 ∨ ¬ m0 ≤ #r mx0 ∨ ¬ m0 ≤ #clean1rows(mx0, i0) ∨ ¬ #clean1rows(mx0, i0) ≤ #r mx0, ∀ n. n < #c mx → ¬ mx[n0 + m0, n] < i ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 10 terms: c abort: F discard: T computed: T

Additional information

Used terms: c

Proof Tree for count1-leq-clean1-ind (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