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Proof step 1 from proof oc-put-toArr-rec (conclusion) (back to tree)

⊦ m < #r mx ∧ n < #c mx ∧ r ≤ #r mx → #oc(i, toArr(mx[m, n, j], r)) = #oc(i, toArr(mx, r)) + (i = j ∧ m < r ⊃ 1;0) - (mx[m, n] = i ∧ m < r ⊃ 1;0)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[m, n, j], r)) ≠ #oc(i, toArr(mx, r)) + (i = j ∧ m < r ⊃ 1;0) - (mx[m, n] = i ∧ m < r ⊃ 1;0), m < #r mx, n < #c mx, r ≤ #r mx ⊦

User applied rule structural induction

And obtained the following new goals

Rule argument

term: r

Additional information

Used simplifier rules: toArr-base: ⊦ toArr(mx, 0) = []; nat freely generated by 0, +1;; ⊦ #oc(a, []) = 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ¬ n < 0; (s)

Proof step 3 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[m, n, j], r0 +1)) ≠ #oc(i, toArr(mx, r0 +1)) + (i = j ∧ m < r0 +1 ⊃ 1;0) - (mx[m, n] = i ∧ m < r0 +1 ⊃ 1;0), n < #c mx, m < #r mx, r0 +1 ≤ #r mx, ∀ i, mx, m, n, j. ¬ ( #oc(i, toArr(mx[m, n, j], r0)) ≠ #oc(i, toArr(mx, r0)) + (i = j ∧ m < r0 ⊃ 1;0) - (mx[m, n] = i ∧ m < r0 ⊃ 1;0) ∧ m < #r mx ∧ n < #c mx ∧ r0 ≤ #r mx) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: toArr-rec: ⊦ toArr(mx, m + 1) = toArr(mx, m) + mx[m]r; ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic 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) ⊦ m < n + 1 ↔ ¬ n < m; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[list-as-arrayinstance-new] ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y); (s)

Proof step 4 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[m, n, j], r0)) + #oc(i, mx[m, n, j][r0]r) ≠ (#oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r)) + (i = j ∧ ¬ r0 < m ⊃ 1;0) - (mx[m, n] = i ∧ ¬ r0 < m ⊃ 1;0), r0 < #r mx, n < #c mx, m < #r mx, ∀ i0, mx0, m0, n0, j0. #oc(i0, toArr(mx0[m0, n0, j0], r0)) = #oc(i0, toArr(mx0, r0)) + (i0 = j0 ∧ m0 < r0 ⊃ 1;0) - (mx0[m0, n0] = i0 ∧ m0 < r0 ⊃ 1;0) ∨ ¬ m0 < #r mx0 ∨ ¬ n0 < #c mx0 ∨ ¬ r0 ≤ #r mx0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:put-selrow, seq:m < #r mx, n < #c mx, r < #r mx ⊦ mx[m, n, j][r]r = (m = r ⊃ mx[r]r[n, j];mx[r]r), subst:[j, n, m, r, mx] → [j, n, m, r0, mx], <all paths>

Additional information

Used terms: j, n, m, r0, mx

Proof step 5 from proof oc-put-toArr-rec (previous step) (back to tree)

m < #r mx, n < #c mx, r < #r mx ⊦ mx[m, n, j][r]r = (m = r ⊃ mx[r]r[n, j];mx[r]r)

Lemma put-selrow

Proof step 6 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[m, n, j], r0)) + #oc(i, (m = r0 ⊃ mx[r0]r[n, j];mx[r0]r)) ≠ (#oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r)) + (i = j ∧ ¬ r0 < m ⊃ 1;0) - (mx[m, n] = i ∧ ¬ r0 < m ⊃ 1;0), m < #r mx, n < #c mx, r0 < #r mx, ∀ i0, mx0, m0, n0, j0. #oc(i0, toArr(mx0[m0, n0, j0], r0)) = #oc(i0, toArr(mx0, r0)) + (i0 = j0 ∧ m0 < r0 ⊃ 1;0) - (mx0[m0, n0] = i0 ∧ m0 < r0 ⊃ 1;0) ∨ ¬ m0 < #r mx0 ∨ ¬ n0 < #c mx0 ∨ ¬ r0 ≤ #r mx0 ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Proof step 7 from proof oc-put-toArr-rec (previous step) (back to tree)

m = r0, #oc(i, toArr(mx[m, n, j], r0)) + #oc(i, mx[r0]r[n, j]) ≠ (#oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r)) + (i = j ∧ ¬ r0 < m ⊃ 1;0) - (mx[m, n] = i ∧ ¬ r0 < m ⊃ 1;0), r0 < #r mx, n < #c mx, m < #r mx, ∀ i0, mx0, m0, n0, j0. #oc(i0, toArr(mx0[m0, n0, j0], r0)) = #oc(i0, toArr(mx0, r0)) + (i0 = j0 ∧ m0 < r0 ⊃ 1;0) - (mx0[m0, n0] = i0 ∧ m0 < r0 ⊃ 1;0) ∨ ¬ m0 < #r mx0 ∨ ¬ n0 < #c mx0 ∨ ¬ r0 ≤ #r mx0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 8 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[r0, n, j], r0)) + #oc(i, mx[r0]r[n, j]) ≠ #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r) + (i = j ⊃ 1;0) - (mx[r0, n] = i ⊃ 1;0), n < #c mx, r0 < #r mx, ∀ i0, mx0, m0, n0, j0. #oc(i0, toArr(mx0[m0, n0, j0], r0)) = #oc(i0, toArr(mx0, r0)) + (i0 = j0 ∧ m0 < r0 ⊃ 1;0) - (mx0[m0, n0] = i0 ∧ m0 < r0 ⊃ 1;0) ∨ ¬ m0 < #r mx0 ∨ ¬ n0 < #c mx0 ∨ ¬ r0 ≤ #r mx0 ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 4 terms: i, mx, r0, n, j abort: F discard: T computed: F

Additional information

Used terms: i, mx, r0, n, j

Proof step 9 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[r0, n, j], r0)) + #oc(i, mx[r0]r[n, j]) ≠ #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r) + (i = j ⊃ 1;0) - (mx[r0, n] = i ⊃ 1;0), r0 < #r mx, n < #c mx, #oc(i, toArr(mx[r0, n, j], r0)) = #oc(i, toArr(mx, r0)) + (i = j ∧ r0 < r0 ⊃ 1;0) - (mx[r0, n] = i ∧ r0 < r0 ⊃ 1;0) ∨ ¬ r0 < #r mx ∨ ¬ n < #c mx ∨ ¬ r0 ≤ #r mx ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) ⊦ m - 0 = m; (s) ⊦ ¬ n < n; (s) 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 oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[r0, n, j], r0)) = #oc(i, toArr(mx, r0)), #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r[n, j]) ≠ #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r) + (i = j ⊃ 1;0) - (mx[r0, n] = i ⊃ 1;0), n < #c mx, r0 < #r mx ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: list-perm[list-as-arrayinstance-new], name:ocs-put-01, seq: ⊦ x[m] ≠ a ∧ a ≠ b ∧ m < # x → #oc(a, x[m, b]) = #oc(a, x), subst:[a, x, m, b] → [i, mx[r0]r, n, j], <all paths>

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s) ⊦ mx[r]r[c] = mx[r, c]; (s) ⊦ 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: i, mx[r0]r, n, j

Proof step 11 from proof oc-put-toArr-rec (previous step) (back to tree)

⊦ x[m] ≠ a ∧ a ≠ b ∧ m < # x → #oc(a, x[m, b]) = #oc(a, x)

A spec-lemma

Proof step 12 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[r0, n, j], r0)) = #oc(i, toArr(mx, r0)), #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r[n, j]) ≠ #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r) + (i = j ⊃ 1;0) - (mx[r0, n] = i ⊃ 1;0), r0 < #r mx, n < #c mx ⊦ mx[r0]r[n] ≠ i ∧ i ≠ j ∧ n < # mx[r0]r

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 13 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[r0, n, j], r0)) = #oc(i, toArr(mx, r0)), #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r[n, j]) ≠ #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r) + (i = j ⊃ 1;0) - (mx[r0, n] = i ⊃ 1;0), n < #c mx, r0 < #r mx ⊦ mx[r0, n] ≠ i ∧ i ≠ j

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: list-perm[list-as-arrayinstance-new], name:ocs-put-02, seq:⊦ x[m] = a ∧ a ≠ b ∧ m < # x → #oc(a, x[m, b]) = #oc(a, x) -1, subst:[a, x, m, b] → [i, mx[r0]r, n, j], <all paths>

Additional information

Used terms: i, mx[r0]r, n, j

Proof step 14 from proof oc-put-toArr-rec (previous step) (back to tree)

⊦ x[m] = a ∧ a ≠ b ∧ m < # x → #oc(a, x[m, b]) = #oc(a, x) -1

A spec-lemma

Proof step 15 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[r0, n, j], r0)) = #oc(i, toArr(mx, r0)), #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r[n, j]) ≠ #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r) + (i = j ⊃ 1;0) - (mx[r0, n] = i ⊃ 1;0), r0 < #r mx, n < #c mx ⊦ mx[r0, n] ≠ i ∧ i ≠ j, mx[r0]r[n] = i ∧ i ≠ j ∧ n < # mx[r0]r

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: getrow-putsame: m < #r mx, n < #c mx ⊦ mx[m]r[n, mx[m, n]] = mx[m]r; 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) ⊦ m - 0 = m; (s) ⊦ m + n - m = n; (s) ⊦ mx[r, c] = a → mx[r, c, a] = mx; (s) ⊦ # mx[r]r = #c mx; (s) ⊦ mx[r]r[c] = mx[r, c]; (s)

Proof step 16 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(j, toArr(mx[r0, n, j], r0)) = #oc(j, toArr(mx, r0)), #oc(j, mx[r0]r[n, j]) ≠ #oc(j, mx[r0]r) + 1, mx[r0, n] ≠ j, n < #c mx, r0 < #r mx ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: list-perm[list-as-arrayinstance-new], name:ocs-put, seq: ⊦ x[m] ≠ a ∧ m < # x → #oc(a, x[m, a]) = #oc(a, x) + 1, subst:[a, x, m] → [j, mx[r0]r, n], <all paths>

Additional information

Used simplifier rules: ⊦ mx[r]r[c] = mx[r, c]; (s) ⊦ # mx[r]r = #c mx; (s) Used terms: j, mx[r0]r, n

Proof step 17 from proof oc-put-toArr-rec (previous step) (back to tree)

⊦ x[m] ≠ a ∧ m < # x → #oc(a, x[m, a]) = #oc(a, x) + 1

A spec-lemma

Proof step 18 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[r0, n, j], r0)) = #oc(i, toArr(mx, r0)), mx[r0]r[n] = i, #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r) -1 ≠ #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r) + (i = j ⊃ 1;0) - (mx[r0, n] = i ⊃ 1;0), i ≠ j, r0 < #r mx, n < #c mx, n < # mx[r0]r ⊦ mx[r0, n] ≠ i ∧ i ≠ j

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) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ n + 0 = n; (s) ⊦ n -1 = n - 1; (s)

Proof step 19 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(mx[r0, n], toArr(mx[r0, n, j], r0)) = #oc(mx[r0, n], toArr(mx, r0)), #oc(mx[r0, n], toArr(mx, r0)) + (#oc(mx[r0, n], mx[r0]r) - 1) ≠ #oc(mx[r0, n], toArr(mx, r0)) + #oc(mx[r0, n], mx[r0]r) - 1, mx[r0, n] ≠ j, r0 < #r mx, n < #c mx ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: nat, name:elim-pred-c, seq:n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1, subst:[n, m] → [#oc(mx[r0, n], toArr(mx, r0)) + #oc(mx[r0, n], mx[r0]r), m]

Additional information

Used simplifier rules: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws) ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x; (s) ⊦ n + 0 = n; (s) ⊦ ¬ a ∈ x → #oc(a, x) = 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: #oc(mx[r0, n], toArr(mx, r0)) + #oc(mx[r0, n], mx[r0]r), m

Proof step 20 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(mx[r0, n], toArr(mx[r0, n, j], r0)) = #oc(mx[r0, n], toArr(mx, r0)), #oc(mx[r0, n], toArr(mx, r0)) + #oc(mx[r0, n], mx[r0]r) = m + 1, #oc(mx[r0, n], toArr(mx, r0)) + #oc(mx[r0, n], mx[r0]r) ≠ 0, #oc(mx[r0, n], toArr(mx, r0)) + (#oc(mx[r0, n], mx[r0]r) - 1) ≠ m, mx[r0, n] ≠ j, n < #c mx, r0 < #r mx ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≠ 0; (s)

Proof step 21 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(mx[r0, n], toArr(mx[r0, n, j], r0)) = #oc(mx[r0, n], toArr(mx, r0)), #oc(mx[r0, n], toArr(mx, r0)) + #oc(mx[r0, n], mx[r0]r) = m + 1, #oc(mx[r0, n], toArr(mx, r0)) + (#oc(mx[r0, n], mx[r0]r) - 1) ≠ m, mx[r0, n] ≠ j, n < #c mx, r0 < #r mx ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat, name:prop-05, seq:m ≤ n ⊦ (n - m) + n0 = n + n0 - m, subst:[n, m, n0] → [#oc(mx[r0, n], mx[r0]r), 1, #oc(mx[r0, n], toArr(mx, r0))], <all paths>

Additional information

Used simplifier rules: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < 1 ↔ m = 0; (s) ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x; (s) ⊦ n + 0 = n; (s) ⊦ ¬ a ∈ x → #oc(a, x) = 0; (s) n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: #oc(mx[r0, n], mx[r0]r), 1, #oc(mx[r0, n], toArr(mx, r0))

Proof step 22 from proof oc-put-toArr-rec (previous step) (back to tree)

m ≤ n ⊦ (n - m) + n0 = n + n0 - m

A spec-lemma

Proof step 23 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(mx[r0, n], toArr(mx[r0, n, j], r0)) = #oc(mx[r0, n], toArr(mx, r0)), #oc(mx[r0, n], toArr(mx, r0)) + #oc(mx[r0, n], mx[r0]r) = m + 1, #oc(mx[r0, n], toArr(mx, r0)) + (#oc(mx[r0, n], mx[r0]r) - 1) ≠ m, mx[r0, n] ≠ j, r0 < #r mx, n < #c mx, ¬ 1 ≤ #oc(mx[r0, n], mx[r0]r) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≠ 0; (s) ⊦ m = m + n ↔ n = 0; (s) ⊦ ¬ a ∈ x → #oc(a, x) = 0; (s) ⊦ n + 0 = n; (s) ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x; (s) ⊦ m < 1 ↔ m = 0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss)

Proof step 24 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(mx[r0, n], toArr(mx, r0)) = m + 1, #oc(mx[r0, n], toArr(mx[r0, n, j], r0)) = m + 1, mx[r0, n] ≠ j, r0 < #r mx, n < #c mx, ¬ mx[r0, n] ∈ mx[r0]r ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: list-del[list-as-arrayinstance-new], name:in-get, seq: ⊦ a ∈ x ↔ (∃ n. n < # x ∧ x[n] = a), subst:[a, x] → [mx[r0, n], mx[r0]r], <all paths>

Additional information

Used terms: mx[r0, n], mx[r0]r

Proof step 25 from proof oc-put-toArr-rec (previous step) (back to tree)

⊦ a ∈ x ↔ (∃ n. n < # x ∧ x[n] = a)

A spec-lemma

Proof step 26 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(mx[r0, n], toArr(mx[r0, n, j], r0)) = m + 1, #oc(mx[r0, n], toArr(mx, r0)) = m + 1, mx[r0, n] ≠ j, n < #c mx, r0 < #r mx, ∀ n0. ¬(n0 < # mx[r0]r ∧ mx[r0]r[n0] = mx[r0, n]) ⊦

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 27 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(mx[r0, n], toArr(mx, r0)) = m + 1, #oc(mx[r0, n], toArr(mx[r0, n, j], r0)) = m + 1, mx[r0, n] ≠ j, r0 < #r mx, n < #c mx, ∀ n0. ¬ n0 < #c mx ∨ mx[r0, n0] ≠ mx[r0, n] ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: n

Additional information

Used terms: n

Proof step 28 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[m, n, j], r0)) + #oc(i, mx[r0]r) ≠ (#oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r)) + (i = j ∧ ¬ r0 < m ⊃ 1;0) - (mx[m, n] = i ∧ ¬ r0 < m ⊃ 1;0), m ≠ r0, r0 < #r mx, n < #c mx, m < #r mx, ∀ i0, mx0, m0, n0, j0. #oc(i0, toArr(mx0[m0, n0, j0], r0)) = #oc(i0, toArr(mx0, r0)) + (i0 = j0 ∧ m0 < r0 ⊃ 1;0) - (mx0[m0, n0] = i0 ∧ m0 < r0 ⊃ 1;0) ∨ ¬ m0 < #r mx0 ∨ ¬ n0 < #c mx0 ∨ ¬ r0 ≤ #r mx0 ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 6 terms: i, mx, m, n, j abort: F discard: T computed: F

Additional information

Used terms: i, mx, m, n, j

Proof step 29 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[m, n, j], r0)) + #oc(i, mx[r0]r) ≠ (#oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r)) + (i = j ∧ ¬ r0 < m ⊃ 1;0) - (mx[m, n] = i ∧ ¬ r0 < m ⊃ 1;0), m ≠ r0, m < #r mx, n < #c mx, r0 < #r mx, #oc(i, toArr(mx[m, n, j], r0)) = #oc(i, toArr(mx, r0)) + (i = j ∧ m < r0 ⊃ 1;0) - (mx[m, n] = i ∧ m < r0 ⊃ 1;0) ∨ ¬ m < #r mx ∨ ¬ n < #c mx ∨ ¬ r0 ≤ #r mx ⊦

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 30 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[m, n, j], r0)) = #oc(i, toArr(mx, r0)) + (i = j ∧ m < r0 ⊃ 1;0) - (mx[m, n] = i ∧ m < r0 ⊃ 1;0), #oc(i, toArr(mx[m, n, j], r0)) + #oc(i, mx[r0]r) ≠ (#oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r)) + (i = j ∧ ¬ r0 < m ⊃ 1;0) - (mx[m, n] = i ∧ ¬ r0 < m ⊃ 1;0), m ≠ r0, r0 < #r mx, n < #c mx, m < #r mx ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Proof step 31 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[m, n, j], r0)) = #oc(i, toArr(mx, r0)) + 1 - (mx[m, n] = i ∧ m < r0 ⊃ 1;0), i = j, #oc(i, toArr(mx[m, n, j], r0)) + #oc(i, mx[r0]r) ≠ (#oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r)) + (i = j ∧ ¬ r0 < m ⊃ 1;0) - (mx[m, n] = i ∧ ¬ r0 < m ⊃ 1;0), m ≠ r0, m < #r mx, n < #c mx, r0 < #r mx, m < r0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 32 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(j, toArr(mx[m, n, j], r0)) = #oc(j, toArr(mx, r0)) + 1 - (mx[m, n] = j ⊃ 1;0), #oc(j, toArr(mx[m, n, j], r0)) + #oc(j, mx[r0]r) ≠ #oc(j, toArr(mx, r0)) + #oc(j, mx[r0]r) + 1 - (mx[m, n] = j ⊃ 1;0), m ≠ r0, m < #r mx, n < #c mx, r0 < #r mx, m < r0, m + 1 < #r mx ⊦

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (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 ⊦ mx[r, c] = a → mx[r, c, a] = mx; (s) ⊦ n + 0 = n; (s)

Proof step 33 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[m, n, j], r0)) = #oc(i, toArr(mx, r0)) + 0 - (mx[m, n] = i ∧ m < r0 ⊃ 1;0), #oc(i, toArr(mx[m, n, j], r0)) + #oc(i, mx[r0]r) ≠ (#oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r)) + (i = j ∧ ¬ r0 < m ⊃ 1;0) - (mx[m, n] = i ∧ ¬ r0 < m ⊃ 1;0), m ≠ r0, m < #r mx, n < #c mx, r0 < #r mx, i ≠ j ∨ ¬ m < r0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

Proof step 34 from proof oc-put-toArr-rec (previous step) (back to tree)

#oc(i, toArr(mx[m, n, j], r0)) = #oc(i, toArr(mx, r0)) - (mx[m, n] = i ⊃ 1;0), #oc(i, toArr(mx[m, n, j], r0)) + #oc(i, mx[r0]r) ≠ #oc(i, toArr(mx, r0)) + #oc(i, mx[r0]r) - (mx[m, n] = i ⊃ 1;0), i ≠ j, m ≠ r0, m < r0, r0 < #r mx, n < #c mx, m < #r mx, m + 1 < #r mx ⊦

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: oc-toArr-positive: m < r, r < #r mx, n < #c mx ⊦ #oc(mx[m, n], toArr(mx, r)) ≠ 0; ⊦ 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 ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < 1 ↔ m = 0; (s) ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws) n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)

Proof Tree for oc-put-toArr-rec (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