Proof step 1 from proof fullcountodd-range (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  countodd(a, m, n, p) ≠ (odd(m) ⊃ (n + 1 - m) / 2;(n - m) / 2), n ≤ # a, m ≤ n, ∀ n0. m ≤ n0 ∧ n0 < n ∧ odd(n0) → p(a[n0]) ⊦

Rule argument

Additional information

Proof step 2 from proof fullcountodd-range (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  countodd(a, m, n, p) ≠ (odd(m) ⊃ ((n - m) + 1) / 2;(n - m) / 2), m ≤ n, n ≤ # a, m ≤ # a, ∀ n0. m ≤ n0 ∧ n0 < n ∧ odd(n0) → p(a[n0]) ⊦

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)

Proof step 3 from proof fullcountodd-range (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 4:

  countodd(a, m, n0 +1, p) ≠ (odd(m) ⊃ ((n0 +1 - m) + 1) / 2;(n0 +1 - m) / 2), m ≤ # a, n0 +1 ≤ # a, m ≤ n0 +1, ∀ n1. m ≤ n1 ∧ n1 < n0 +1 ∧ odd(n1) → p(a[n1]), ∀ a, m, p. ¬ ( countodd(a, m, n0, p) ≠ (odd(m) ⊃ ((n0 - m) + 1) / 2;(n0 - m) / 2) ∧ m ≤ n0 ∧ n0 ≤ # a ∧ m ≤ # a ∧ (∀ n1. m ≤ n1 ∧ n1 < n0 ∧ odd(n1) → p(a[n1]))) ⊦

Rule argument

Additional information

Used simplifier rules: countodd-base: m ≤ # a ⊦ countodd(a, m, m, p) = 0; nat freely generated by 0, +1;; ⊦ m ≤ 0 ↔ m = 0; (s) ⊦ ¬ n < 0; (s) ⊦ 0 ≤ m; (s) ⊦ ¬ odd(0); (s)

Proof step 4 from proof fullcountodd-range (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  a ≠ [], countodd(a, m, n0 + 1, p) ≠ (odd(m) ⊃ ((n0 + 1 - m) + 1) / 2;(n0 + 1 - m) / 2), n0 < # a, m ≤ n0 + 1, m ≤ # a, ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]), ∀ a0, m0, p0. countodd(a0, m0, n0, p0) = (odd(m0) ⊃ ((n0 - m0) + 1) / 2;(n0 - m0) / 2) ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # a0 ∨ ¬ m0 ≤ # a0 ∨ ¬(∀ n1. m0 ≤ n1 ∧ n1 < n0 ∧ odd(n1) → p0(a0[n1])) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (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 < n + 1 ↔ ¬ n < m; (s) ⊦ m +1 = m + 1; (s) ⊦ m + 1 ≤ n ↔ m < n; (s)

Proof step 5 from proof fullcountodd-range (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

  m ≤ n, n ≤ # a ⊦ countodd(a, m, n + 1, p) = countodd(a, m, n, p) + (odd(n) ∧ p(a[n]) ⊃ 1;0)

• Proof step 7:

  countodd(a, m, n0, p) + (odd(n0) ∧ p(a[n0]) ⊃ 1;0) ≠ (odd(m) ⊃ ((n0 + 1 - m) + 1) / 2;(n0 + 1 - m) / 2), a ≠ [], n0 < # a, m ≤ # a, m ≤ n0 + 1, m ≤ n0, n0 ≤ # a, ∀ a0, m0, p0. countodd(a0, m0, n0, p0) = (odd(m0) ⊃ ((n0 - m0) + 1) / 2;(n0 - m0) / 2) ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # a0 ∨ ¬ m0 ≤ # a0 ∨ ¬(∀ n1. m0 ≤ n1 ∧ n1 < n0 ∧ odd(n1) → p0(a0[n1])), ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Used simplifier rules: countodd-base: m ≤ # a ⊦ countodd(a, m, m, p) = 0; ⊦ m < n → m ≤ n; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s) eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m + 1 ≠ 0; (s) ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → (n ≤ m ↔ false); (s) ⊦ odd(n + 1) ↔ even(n); (s) ⊦ m + n - n + n1 = m - n1; (s) ⊦ m - m = 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: a, m, n0, p

Proof step 6 from proof fullcountodd-range (previous step) (back to tree)

Lemma countodd-rec

Proof step 7 from proof fullcountodd-range (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  countodd(a, m, n0, p) + (odd(n0) ∧ p(a[n0]) ⊃ 1;0) ≠ (odd(m) ⊃ (n0 - m) / 2 + 1;((n0 - m) + 1) / 2), a ≠ [], m < # a, n0 < # a, m ≤ n0 + 1, m ≤ n0, ∀ a0, m0, p0. countodd(a0, m0, n0, p0) = (odd(m0) ⊃ ((n0 - m0) + 1) / 2;(n0 - m0) / 2) ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ # a0 ∨ ¬ m0 ≤ # a0 ∨ ¬(∀ n1. m0 ≤ n1 ∧ n1 < n0 ∧ odd(n1) → p0(a0[n1])), ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) fle-01: ⊦ 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 ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ (m + 2) / 2 = m / 2 + 1; (s)

Proof step 8 from proof fullcountodd-range (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 9:

  countodd(a, m, n0, p) + (odd(n0) ∧ p(a[n0]) ⊃ 1;0) ≠ (odd(m) ⊃ (n0 - m) / 2 + 1;((n0 - m) + 1) / 2), a ≠ [], n0 < # a, m < # a, m ≤ n0, m ≤ n0 + 1, ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]), countodd(a, m, n0, p) = (odd(m) ⊃ ((n0 - m) + 1) / 2;(n0 - m) / 2) ∨ ¬ m ≤ n0 ∨ ¬ n0 ≤ # a ∨ ¬ m ≤ # a ∨ ¬(∀ n1. m ≤ n1 ∧ n1 < n0 ∧ odd(n1) → p(a[n1])) ⊦

Rule argument

Additional information

Used terms: a, m, p

Proof step 9 from proof fullcountodd-range (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  countodd(a, m, n0, p) + (odd(n0) ∧ p(a[n0]) ⊃ 1;0) ≠ (odd(m) ⊃ (n0 - m) / 2 + 1;((n0 - m) + 1) / 2), a ≠ [], m < # a, n0 < # a, m ≤ n0 + 1, m ≤ n0, ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]), countodd(a, m, n0, p) = (odd(m) ⊃ ((n0 - m) + 1) / 2;(n0 - m) / 2) ∨ ¬(m ≤ n1 ∧ n1 < n0 ∧ odd(n1) → p(a[n1])) ⊦

Rule argument

Additional information

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

Proof step 10 from proof fullcountodd-range (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 11:

  countodd(a, m, n0, p) + (odd(n0) ∧ p(a[n0]) ⊃ 1;0) ≠ (odd(m) ⊃ (n0 - m) / 2 + 1;((n0 - m) + 1) / 2), a ≠ [], n0 < # a, m < # a, m ≤ n0, m ≤ n0 + 1, m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]), countodd(a, m, n0, p) = (odd(m) ⊃ ((n0 - m) + 1) / 2;(n0 - m) / 2) ∨ ¬(m ≤ n1 ∧ n1 < n0 ∧ odd(n1) → p(a[n1])) ⊦

Rule argument

Additional information

Used terms: n1

Proof step 11 from proof fullcountodd-range (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 12:

  countodd(a, m, n0, p) = (odd(m) ⊃ ((n0 - m) + 1) / 2;(n0 - m) / 2), countodd(a, m, n0, p) + (odd(n0) ∧ p(a[n0]) ⊃ 1;0) ≠ (odd(m) ⊃ (n0 - m) / 2 + 1;((n0 - m) + 1) / 2), a ≠ [], m < # a, n0 < # a, m ≤ n0 + 1, m ≤ n0, m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

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

Proof step 12 from proof fullcountodd-range (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 13:

  countodd(a, m, n0, p) = ((n0 - m) + 1) / 2, countodd(a, m, n0, p) + (odd(n0) ∧ p(a[n0]) ⊃ 1;0) ≠ (odd(m) ⊃ (n0 - m) / 2 + 1;((n0 - m) + 1) / 2), a ≠ [], n0 < # a, m < # a, odd(m), m ≤ n0, m ≤ n0 + 1, m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]) ⊦

• Proof step 22:

  countodd(a, m, n0, p) = (n0 - m) / 2, countodd(a, m, n0, p) + (odd(n0) ∧ p(a[n0]) ⊃ 1;0) ≠ (odd(m) ⊃ (n0 - m) / 2 + 1;((n0 - m) + 1) / 2), a ≠ [], n0 < # a, m < # a, m ≤ n0, m ≤ n0 + 1, ¬ odd(m), m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Proof step 13 from proof fullcountodd-range (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  countodd(a, m, n0, p) = ((n0 - m) + 1) / 2, countodd(a, m, n0, p) + (odd(n0) ∧ p(a[n0]) ⊃ 1;0) ≠ (n0 - m) / 2 + 1, a ≠ [], m < # a, n0 < # a, odd(m), m ≤ n0 + 1, m ≤ n0, m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Proof step 14 from proof fullcountodd-range (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 15:

  countodd(a, m, n0, p) = (n2 + 1) / 2, n0 = n2 + m, countodd(a, m, n0, p) + (odd(n0) ∧ p(a[n0]) ⊃ 1;0) ≠ n2 / 2 + 1, a ≠ [], n0 < # a, m < # a, odd(m), m ≤ n0, m ≤ n0 + 1, m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0, m, n2

Proof step 15 from proof fullcountodd-range (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  countodd(a, m, n2 + m, p) = (n2 + 1) / 2, countodd(a, m, n2 + m, p) + (odd(n2 + m) ∧ p(a[n2 + m]) ⊃ 1;0) ≠ n2 / 2 + 1, a ≠ [], m < # a, n2 + m < # a, odd(m), m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

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

Proof step 16 from proof fullcountodd-range (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 17:

  ⊦ odd(m + n) ↔ odd(m) ∧ even(n) ∨ even(m) ∧ odd(n)

• Proof step 18:

  countodd(a, m, n2 + m, p) = (n2 + 1) / 2, countodd(a, m, n2 + m, p) + ((odd(n2) ∧ even(m) ∨ even(n2) ∧ odd(m)) ∧ p(a[n2 + m]) ⊃ 1;0) ≠ n2 / 2 + 1, a ≠ [], n2 + m < # a, m < # a, odd(m), m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Used terms: n2, m

Proof step 17 from proof fullcountodd-range (previous step) (back to tree)

A spec-lemma

Proof step 18 from proof fullcountodd-range (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  countodd(a, m, n2 + m, p) = (n2 + 1) / 2, countodd(a, m, n2 + m, p) + (even(n2) ∧ p(a[n2 + m]) ⊃ 1;0) ≠ n2 / 2 + 1, a ≠ [], m < # a, n2 + m < # a, odd(m), m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ odd(n) → ¬ even(n); (s)

Proof step 19 from proof fullcountodd-range (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 20:

  countodd(a, m, n2 + m, p) = (n2 + 1) / 2, countodd(a, m, n2 + m, p) + 0 ≠ n2 / 2 + 1, a ≠ [], n2 + m < # a, m < # a, odd(m), m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]), ¬ even(n2) ∨ ¬ p(a[n2 + m]) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ even(n) → (n + 1) / 2 = n / 2; (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

Proof step 20 from proof fullcountodd-range (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  countodd(a, m, n2 + m, p) = n2 / 2, a ≠ [], m < # a, n2 + m < # a, even(n2), odd(m), ¬ p(a[n2 + m]), m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m = m + n ↔ n = 0; (s) ⊦ even(n) → (n + 1) / 2 = n / 2; (s) ⊦ odd(n) → (n + 1) / 2 = n / 2 + 1; (s) ⊦ even(n) ↔ ¬ odd(n); (ss) ⊦ n + 0 = n; (s)

Proof step 21 from proof fullcountodd-range (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m + n; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ ¬ n < n; (s) ⊦ odd(n) ↔ ¬ even(n); (ss) ⊦ even(m) → (even(m + n) ↔ even(n)); (s) ⊦ odd(n) → ¬ even(n); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: n2 + m

Proof step 22 from proof fullcountodd-range (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  countodd(a, m, n0, p) = (n0 - m) / 2, countodd(a, m, n0, p) + (odd(n0) ∧ p(a[n0]) ⊃ 1;0) ≠ ((n0 - m) + 1) / 2, a ≠ [], m < # a, n0 < # a, even(m), m ≤ n0 + 1, m ≤ n0, m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ odd(n) ↔ ¬ even(n); (ss)

Proof step 23 from proof fullcountodd-range (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 24:

  countodd(a, m, n0, p) = n2 / 2, n0 = n2 + m, countodd(a, m, n0, p) + (odd(n0) ∧ p(a[n0]) ⊃ 1;0) ≠ (n2 + 1) / 2, a ≠ [], n0 < # a, m < # a, even(m), m ≤ n0, m ≤ n0 + 1, m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n0 < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0, m, n2

Proof step 24 from proof fullcountodd-range (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  countodd(a, m, n2 + m, p) = n2 / 2, n2 / 2 + (odd(n2 + m) ∧ p(a[n2 + m]) ⊃ 1;0) ≠ (n2 + 1) / 2, a ≠ [], m < # a, n2 + m < # a, even(m), m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

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

Proof step 25 from proof fullcountodd-range (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  ⊦ odd(m + n) ↔ odd(m) ∧ even(n) ∨ even(m) ∧ odd(n)

• Proof step 27:

  countodd(a, m, n2 + m, p) = n2 / 2, n2 / 2 + ((odd(n2) ∧ even(m) ∨ even(n2) ∧ odd(m)) ∧ p(a[n2 + m]) ⊃ 1;0) ≠ (n2 + 1) / 2, a ≠ [], n2 + m < # a, m < # a, even(m), m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Used terms: n2, m

Proof step 26 from proof fullcountodd-range (previous step) (back to tree)

A spec-lemma

Proof step 27 from proof fullcountodd-range (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  countodd(a, m, n2 + m, p) = n2 / 2, n2 / 2 + (odd(n2) ∧ p(a[n2 + m]) ⊃ 1;0) ≠ (n2 + 1) / 2, a ≠ [], m < # a, n2 + m < # a, even(m), m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ even(n) → ¬ odd(n); (s)

Proof step 28 from proof fullcountodd-range (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 29:

  countodd(a, m, n2 + m, p) = n2 / 2, n2 / 2 + 0 ≠ (n2 + 1) / 2, a ≠ [], n2 + m < # a, m < # a, even(m), m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]), ¬ odd(n2) ∨ ¬ p(a[n2 + m]) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ odd(n) → (n + 1) / 2 = n / 2 + 1; (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

Proof step 29 from proof fullcountodd-range (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  countodd(a, m, n2 + m, p) = n2 / 2, a ≠ [], n2 + m < # a, m < # a, even(m), odd(n2), ¬ p(a[n2 + m]), m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]), ∀ n1. m ≤ n1 ∧ ¬ n2 + m < n1 ∧ odd(n1) → p(a[n1]) ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m = m + n ↔ n = 0; (s) ⊦ odd(n) → (n + 1) / 2 = n / 2 + 1; (s) ⊦ even(n) → (n + 1) / 2 = n / 2; (s) ⊦ odd(n) ↔ ¬ even(n); (ss) ⊦ n + 0 = n; (s)

Proof step 30 from proof fullcountodd-range (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m + n; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ ¬ n < n; (s) ⊦ odd(n) ↔ ¬ even(n); (ss) ⊦ even(m) → (even(m + n) ↔ even(n)); (s) ⊦ odd(n) → ¬ even(n); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: n2 + m