Proof step 1 from proof diff-size (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(#(h \ rs) ≤ #((h -- r) \ rs) ↔ false), r ∈ h \ rs ⊦

Rule argument

Additional information

Proof step 2 from proof diff-size (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 3:

  ¬(#(h \ (rs0 ++ r0)) ≤ #((h -- r) \ (rs0 ++ r0)) ↔ false), r ∈ h \ (rs0 ++ r0), ∀ h, r. ¬(¬(#(h \ rs0) ≤ #((h -- r) \ rs0) ↔ false) ∧ r ∈ h \ rs0) ⊦

Rule argument

Additional information

Used simplifier rules: diffs-emp: ⊦ h0 \ ∅ = h0; Set[A] generated by ∅, ++; ⊦ a ∈ st → #(st -- a) = (# st).-1; (s) ⊦ n.-1 = n - 1; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < 1 ↔ m = 0; (s) ⊦ # st = 0 ↔ st = ∅; (s) ⊦ ¬ a ∈ ∅; (s) n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s) ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ ¬ n < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

Proof step 3 from proof diff-size (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 4:

  r ≠ r0, #((h -- r0) \ rs0) ≤ #((h -- r -- r0) \ rs0), r ∈ h, ¬ r ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

Rule argument

Additional information

Used simplifier rules: diffs-ins: ⊦ h0 \ (rs ++ r) = (h0 -- r) \ rs; in-diffs: ⊦ r ∈ h \ rs ↔ r ∈ h ∧ ¬ r ∈ rs; ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; (s)

Proof step 4 from proof diff-size (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 5:

  ⊦ (h -- r) \ rs = (h \ rs) -- r

• Proof step 6:

  r ≠ r0, #((h \ rs0) -- r0) ≤ #((h -- r -- r0) \ rs0), r ∈ h, ¬ r ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

Rule argument

Additional information

Used terms: h, r0, rs0

Proof step 5 from proof diff-size (previous step) (back to tree)

Lemma diffs-sw

Proof step 6 from proof diff-size (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  r ≠ r0, #((h \ rs0) -- r0) ≤ #((h -- r -- r0) \ rs0), r ∈ h, ¬ r ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

Rule argument

Additional information

Proof step 7 from proof diff-size (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 8:

  ⊦ (h -- r) \ rs = (h \ rs) -- r

• Proof step 9:

  r ≠ r0, #((h \ rs0) -- r0) ≤ #(((h -- r) \ rs0) -- r0), r ∈ h, ¬ r ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

Rule argument

Additional information

Used terms: h -- r, r0, rs0

Proof step 8 from proof diff-size (previous step) (back to tree)

Lemma diffs-sw

Proof step 9 from proof diff-size (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

  ⊦ r ∈ h → #(h -- r) = # h - 1

• Proof step 11:

  r ≠ r0, #((h \ rs0) -- r0) ≤ #(((h -- r) \ rs0) -- r0), r ∈ h, ¬ r ∈ rs0, ¬ r0 ∈ h \ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

• Proof step 16:

  r ≠ r0, #(h \ rs0) - 1 ≤ #(((h -- r) \ rs0) -- r0), r ∈ h, r0 ∈ h \ rs0, ¬ r ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

Rule argument

Additional information

Used terms: h \ rs0, r0

Proof step 10 from proof diff-size (previous step) (back to tree)

Lemma rm-size

Proof step 11 from proof diff-size (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  r ≠ r0, #(h \ rs0) ≤ #(((h -- r) \ rs0) -- r0), r ∈ h, ¬ r ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦ r0 ∈ h ∧ ¬ r0 ∈ rs0

Rule argument

Additional information

Used simplifier rules: in-diffs: ⊦ r ∈ h \ rs ↔ r ∈ h ∧ ¬ r ∈ rs; ⊦ ¬ a ∈ st → st -- a = st; (s)

Proof step 12 from proof diff-size (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

  ⊦ ¬ a ∈ st → st -- a = st

• Proof step 14:

  r ≠ r0, #(h \ rs0) ≤ #((h -- r) \ rs0), r ∈ h, ¬ r ∈ rs0, ¬ r0 ∈ (h -- r) \ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦ r0 ∈ h ∧ ¬ r0 ∈ rs0

Rule argument

Additional information

Used simplifier rules: in-diffs: ⊦ r ∈ h \ rs ↔ r ∈ h ∧ ¬ r ∈ rs; ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; (s) Used terms: (h -- r) \ rs0, r0

Proof step 13 from proof diff-size (previous step) (back to tree)

A spec-lemma

Proof step 14 from proof diff-size (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  r ≠ r0, #(h \ rs0) ≤ #((h -- r) \ rs0), r ∈ h, ¬ r ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦ r0 ∈ h ∧ ¬ r0 ∈ rs0

Rule argument

Additional information

Used simplifier rules: in-diffs: ⊦ r ∈ h \ rs ↔ r ∈ h ∧ ¬ r ∈ rs; ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; (s)

Proof step 15 from proof diff-size (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: h, r

Proof step 16 from proof diff-size (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 17:

  ⊦ r ∈ h → #(h -- r) = # h - 1

• Proof step 18:

  r ≠ r0, #(h \ rs0) - 1 ≤ #((h -- r) \ rs0) - 1, r ∈ h, r0 ∈ h \ rs0, r0 ∈ (h -- r) \ rs0, ¬ r ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

Rule argument

Additional information

Used simplifier rules: in-diffs: ⊦ r ∈ h \ rs ↔ r ∈ h ∧ ¬ r ∈ rs; ⊦ ¬ a ∈ st → st -- a = st; (s) ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; (s) Used terms: (h -- r) \ rs0, r0

Proof step 17 from proof diff-size (previous step) (back to tree)

Lemma rm-size

Proof step 18 from proof diff-size (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 19:

  #(h \ rs0) = 0, r ≠ r0, #(h \ rs0) - 1 ≤ #((h -- r) \ rs0) - 1, r ∈ h, r0 ∈ (h -- r) \ rs0, r0 ∈ h \ rs0, ¬ r ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

• Proof step 21:

  #(h \ rs0) = n0 + 1, #(h \ rs0) ≠ 0, r ≠ r0, n0 ≤ #((h -- r) \ rs0) - 1, r ∈ h, r0 ∈ (h -- r) \ rs0, r0 ∈ h \ rs0, ¬ r ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

Rule argument

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: #(h \ rs0), n0

Proof step 19 from proof diff-size (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  h \ rs0 = ∅, r ≠ r0, 0 - 1 ≤ #((h -- r) \ rs0) - 1, r ∈ h, r0 ∈ h, ¬ r ∈ rs0, ¬ r0 ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

Rule argument

Additional information

Used simplifier rules: in-diffs: ⊦ r ∈ h \ rs ↔ r ∈ h ∧ ¬ r ∈ rs; ⊦ # ∅ = 0; (s) ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; (s) ⊦ # st = 0 ↔ st = ∅; (s)

Proof step 20 from proof diff-size (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ # ∅ = 0; (s) Used terms: h, r

Proof step 21 from proof diff-size (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  #(h \ rs0) = n0 + 1, h \ rs0 ≠ ∅, r ≠ r0, n0 ≤ #((h -- r) \ rs0) - 1, r ∈ h, r0 ∈ h, ¬ r ∈ rs0, ¬ r0 ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

Rule argument

Additional information

Used simplifier rules: in-diffs: ⊦ r ∈ h \ rs ↔ r ∈ h ∧ ¬ r ∈ rs; ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; (s) ⊦ # st = 0 ↔ st = ∅; (s)

Proof step 22 from proof diff-size (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 23:

  #((h -- r) \ rs0) = 0, #(h \ rs0) = n0 + 1, h \ rs0 ≠ ∅, r ≠ r0, n0 ≤ #((h -- r) \ rs0) - 1, r ∈ h, r0 ∈ h, ¬ r ∈ rs0, ¬ r0 ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

• Proof step 25:

  #(h \ rs0) = n0 + 1, #((h -- r) \ rs0) = n1 + 1, #((h -- r) \ rs0) ≠ 0, h \ rs0 ≠ ∅, r ≠ r0, n0 ≤ n1, r ∈ h, r0 ∈ h, ¬ r ∈ rs0, ¬ r0 ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

Rule argument

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: #((h -- r) \ rs0), n1

Proof step 23 from proof diff-size (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  #(h \ rs0) = n0 + 1, (h -- r) \ rs0 = ∅, h \ rs0 ≠ ∅, r ≠ r0, n0 ≤ 0 - 1, r ∈ h, r0 ∈ h, ¬ r ∈ rs0, ¬ r0 ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # ∅ = 0; (s) ⊦ # st = 0 ↔ st = ∅; (s)

Proof step 24 from proof diff-size (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ # ∅ = 0; (s) Used terms: h -- r, r0

Proof step 25 from proof diff-size (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  #((h -- r) \ rs0) = n1 + 1, #(h \ rs0) = n0 + 1, h \ rs0 ≠ ∅, (h -- r) \ rs0 ≠ ∅, r ≠ r0, n0 ≤ n1, r ∈ h, r0 ∈ h, ¬ r ∈ rs0, ¬ r0 ∈ rs0, ∀ h0, r1. ¬ #(h0 \ rs0) ≤ #((h0 -- r1) \ rs0) ∨ ¬ r1 ∈ h0 ∨ r1 ∈ rs0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # st = 0 ↔ st = ∅; (s)

Proof step 26 from proof diff-size (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic 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 ⊦ m + 1 ≠ 0; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m ≤ m0 ∧ n ≤ n0 → ¬ m0 + n0 < m + n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: h, r