Proof step 1 from proof difference (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  count(a, bag0 \ bag1) ≠ (count(a, bag1) ≤ count(a, bag0) ⊃ count(a, bag0) - count(a, bag1);0) ⊦

Rule argument

Additional information

Proof step 2 from proof difference (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 3:

  count(a, bag0 \ bag1) = (count(a, bag1) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag1);0), count(a, bag0 \ bag1) ≠ (count(a, bag1) ≤ count(a, bag0) ⊃ count(a, bag0) - count(a, bag1);0) ⊦

• Proof step 4:

  count(a, bag0 \ bag1) ≠ (count(a, bag1) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag1);0), count(a, bag0 \ bag1) ≠ (count(a, bag1) ≤ count(a, bag0) ⊃ count(a, bag0) - count(a, bag1);0) ⊦

Rule argument

Additional information

Used formulas: count(a, bag0 \ bag1) = (count(a, bag1) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag1);0)

Proof step 3 from proof difference (previous step) (back to tree)

User applied rule case distinction

And closed the goal!

Rule argument

Additional information

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

Proof step 4 from proof difference (previous step) (back to tree)

User applied rule weakening

And obtained the following new goals

• Proof step 5:

  count(a, bag0 \ bag1) ≠ (count(a, bag1) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag1);0) ⊦

Rule argument

Additional information

Proof step 5 from proof difference (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 6:

  count(a, bag0 \ ∅) ≠ (count(a, ∅) < count(a, bag0) ⊃ count(a, bag0) - count(a, ∅);0) ⊦

• Proof step 8:

  count(a, bag0 \ (bag ++ a0)) ≠ (count(a, bag ++ a0) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag ++ a0);0), ∀ a, bag0. count(a, bag0 \ bag) = (count(a, bag) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag);0) ⊦

Rule argument

Additional information

Used simplifier rules: bag generated by ∅, ++;;

Proof step 6 from proof difference (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  count(a, bag0) ≠ (count(a, bag0) ≠ 0 ⊃ count(a, bag0);0) ⊦

Rule argument

Additional information

Used simplifier rules: diff-empty: ⊦ bag0 \ ∅ = bag0; ⊦ m - 0 = m; (s) ⊦ count(a, ∅) = 0; (s) ⊦ 0 < m ↔ m ≠ 0; (s)

Proof step 7 from proof difference (previous step) (back to tree)

User applied rule case distinction

And closed the goal!

Rule argument

Additional information

Proof step 8 from proof difference (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  count(a, (bag0 -- a0) \ bag) ≠ (count(a, bag ++ a0) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag ++ a0);0), ∀ a, bag0. count(a, bag0 \ bag) = (count(a, bag) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag);0) ⊦

Rule argument

Additional information

Used simplifier rules: diff-rec: ⊦ bag0 \ (bag1 ++ a) = (bag0 -- a) \ bag1;

Proof step 9 from proof difference (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 10:

  count(a, (bag0 -- a0) \ bag) = (count(a, bag) < count(a, bag0 -- a0) ⊃ count(a, bag0 -- a0) - count(a, bag);0), count(a, (bag0 -- a0) \ bag) ≠ (count(a, bag ++ a0) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag ++ a0);0) ⊦

Rule argument

Additional information

Used terms: a, bag0 -- a0

Proof step 10 from proof difference (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  count(a, (bag0 -- a0) \ bag) = (count(a, bag) < count(a, bag0) - (a = a0 ∧ count(a0, bag0) ≠ 0 ⊃ 1;0) ⊃ count(a, bag0) - (a = a0 ∧ count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a, bag);0), count(a, (bag0 -- a0) \ bag) ≠ (count(a, bag ++ a0) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag ++ a0);0) ⊦

Rule argument

Additional information

Used simplifier rules: count-del: ⊦ count(a, bag -- b) = count(a, bag) - (a = b ∧ count(a, bag) ≠ 0 ⊃ 1;0);

Proof step 11 from proof difference (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 12:

  ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag)

• Proof step 13:

  a = a0, count(a, (bag0 -- a0) \ bag) = (count(a, bag) < count(a, bag0) - (a = a0 ∧ count(a0, bag0) ≠ 0 ⊃ 1;0) ⊃ count(a, bag0) - (a = a0 ∧ count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a, bag);0), count(a, (bag0 -- a0) \ bag) ≠ (count(a, bag ++ a0) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag ++ a0);0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m - 0 = m; (s) Used terms: a0, bag, a

Proof step 12 from proof difference (previous step) (back to tree)

A spec-lemma

Proof step 13 from proof difference (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  count(a0, (bag0 -- a0) \ bag) = (count(a0, bag) < count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) ⊃ count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a0, bag);0), count(a0, (bag0 -- a0) \ bag) ≠ (count(a0, bag) + 1 < count(a0, bag0) ⊃ count(a0, bag0) - count(a0, bag) + 1;0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ count(a, bag ++ a) = count(a, bag) + 1; (s)

Proof step 14 from proof difference (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 15:

  count(a0, (bag0 -- a0) \ bag) = (count(a0, bag) < count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) ⊃ count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a0, bag);0), count(a0, (bag0 -- a0) \ bag) ≠ count(a0, bag0) - count(a0, bag) + 1, count(a0, bag) + 1 < count(a0, bag0) ⊦

• Proof step 17:

  count(a0, (bag0 -- a0) \ bag) = (count(a0, bag) < count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) ⊃ count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a0, bag);0), count(a0, (bag0 -- a0) \ bag) ≠ 0, ¬ count(a0, bag) + 1 < count(a0, bag0) ⊦

Rule argument

Additional information

Proof step 15 from proof difference (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  count(a0, (bag0 -- a0) \ bag) = count(a0, bag0) - 1 - count(a0, bag), count(a0, bag0) - 1 - count(a0, bag) ≠ count(a0, bag0) - count(a0, bag) + 1, count(a0, bag) + 1 < count(a0, bag0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; (s) ⊦ m < 1 ↔ m = 0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss)

Proof step 16 from proof difference (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m + 1 ≠ 0; (s) ⊦ m + n - n + n1 = m - n1; (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: count(a0, bag0), m

Proof step 17 from proof difference (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  count(a0, (bag0 -- a0) \ bag) = (count(a0, bag) < count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) ⊃ count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a0, bag);0), count(a0, (bag0 -- a0) \ bag) ≠ 0, count(a0, bag0) ≤ count(a0, bag) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 18 from proof difference (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 19:

  count(a0, (bag0 -- a0) \ bag) = count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a0, bag), count(a0, (bag0 -- a0) \ bag) ≠ 0, count(a0, bag) < count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0), count(a0, bag0) ≤ count(a0, bag) + 1 ⊦

Rule argument

Additional information

Proof step 19 from proof difference (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  count(a0, (bag0 -- a0) \ bag) = count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a0, bag), count(a0, (bag0 -- a0) \ bag) ≠ 0, count(a0, bag) < count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0), count(a0, bag0) ≤ count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0), count(a0, bag0) ≤ count(a0, bag) + 1 ⊦

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 le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 20 from proof difference (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 21:

  count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) = n0 + count(a0, bag), count(a0, (bag0 -- a0) \ bag) = n0, count(a0, (bag0 -- a0) \ bag) ≠ 0, count(a0, bag) < count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0), count(a0, bag0) ≤ count(a0, bag) + 1, count(a0, bag0) ≤ count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0), count(a0, bag) ≤ count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0), count(a0, bag), n0

Proof step 21 from proof difference (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) = count(a0, (bag0 -- a0) \ bag) + count(a0, bag), count(a0, (bag0 -- a0) \ bag) ≠ 0, count(a0, bag0) ≤ count(a0, (bag0 -- a0) \ bag) + count(a0, bag), count(a0, bag0) ≤ count(a0, bag) + 1 ⊦

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) ⊦ m < n → n ≠ 0; (s) le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n → m ≤ n; (s)

Proof step 22 from proof difference (previous step) (back to tree)

Heuristic cut applied rule cut formula

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del-notin: ⊦ count(a, bag) = 0 → bag -- a = bag; leprf: ⊦ m ≤ m - 1 → m = 0; (forward) from specification nat ⊦ m - 0 = m; (s) let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used formulas: count(a0, bag0) = 0