Proof step 1 from proof writersOk-prop-FO (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  isPropFO(n, l, bq(t)), writersOk(bq, pq, writers), ¬ writersOk(bq(t := bq(t) -p n), pq(l := pq(l) + FO(t) '), writers) ⊦

Rule argument

Additional information

Proof step 2 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  isPropFO(n, l, bq(t)), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦ (t0 ∈ writers(l0) → l1 ≠ l0 → ¬ t0 ∈ writers(l1)) ∧ (¬ t0 ∈ writers(l0) → (l = l0 ⊃ pq(l0) + FO(t) ';pq(l0)) |W | t0 = [] ∧ (t = t0 ⊃ bq(t0) -p n;bq(t0)) |WL l0 = [])

Rule argument

Additional information

Used simplifier rules: writersOk: ⊦ writersOk(bq, p, writers) ↔ (∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → p(l) |W | t = [] ∧ bq(t) |WL l = []));

Proof step 3 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 4:

  l1 ≠ l0, isPropFO(n, l, bq(t)), t0 ∈ writers(l1), t0 ∈ writers(l0), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦

• Proof step 8:

  isPropFO(n, l, bq(t)), ¬ t0 ∈ writers(l0), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦ (l = l0 ⊃ pq(l0) + FO(t) ';pq(l0)) |W | t0 = [] ∧ (t = t0 ⊃ bq(t0) -p n;bq(t0)) |WL l0 = []

Rule argument

Additional information

Proof step 4 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  # writers(l1) ≠ 0, # writers(l0) ≠ 0, l1 ≠ l0, writers(l1) ≠ ∅, writers(l0) ≠ ∅, isPropFO(n, l, bq(t)), t0 ∈ writers(l1), t0 ∈ writers(l0), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic

Proof step 5 from proof writersOk-prop-FO (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 6:

  # writers(l1) ≠ 0, # writers(l0) ≠ 0, l1 ≠ l0, writers(l1) ≠ ∅, writers(l0) ≠ ∅, isPropFO(n, l, bq(t)), t0 ∈ writers(l0), t0 ∈ writers(l1), t0 ∈ writers(l0) → (∀ l1. l1 ≠ l0 → ¬ t0 ∈ writers(l1)), ¬ t0 ∈ writers(l0) → pq(l0) |W | t0 = [] ∧ bq(t0) |WL l0 = [], ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦

Rule argument

Additional information

Used terms: l0, t0

Proof step 6 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  # writers(l1) ≠ 0, # writers(l0) ≠ 0, l1 ≠ l0, writers(l1) ≠ ∅, writers(l0) ≠ ∅, isPropFO(n, l, bq(t)), t0 ∈ writers(l1), t0 ∈ writers(l0), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []), ∀ l1. l1 ≠ l0 → ¬ t0 ∈ writers(l1) ⊦

Rule argument

Additional information

Proof step 7 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: l1

Proof step 8 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 9:

  (l = l0 ⊃ pq(l0) + FO(t) ';pq(l0)) |W | t0 ≠ [], isPropFO(n, l, bq(t)), ¬ t0 ∈ writers(l0), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦

• Proof step 13:

  (t = t0 ⊃ bq(t0) -p n;bq(t0)) |WL l0 ≠ [], isPropFO(n, l, bq(t)), ¬ t0 ∈ writers(l0), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦

Rule argument

Additional information

Proof step 9 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 10:

  l = l0, (pq(l0) + FO(t) ')|W | t0 ≠ [], isPropFO(n, l, bq(t)), ¬ t0 ∈ writers(l0), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦

• Proof step 12:

  l ≠ l0, pq(l0) |W | t0 ≠ [], isPropFO(n, l, bq(t)), ¬ t0 ∈ writers(l0), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦

Rule argument

Additional information

Proof step 10 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  pq(l0) |W | t0 ≠ [], isPropFO(n, l0, bq(t)), ¬ t0 ∈ writers(l0), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ x + [] = x; (s) ⊦ FO(t) ' |W = []; (s) ⊦ (fowl + fowl0)|W = fowl |W + fowl0 |W; (s)

Proof step 11 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: l0, t0

Proof step 12 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: l0, t0

Proof step 13 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 14:

  t = t0, bq(t0) -p n |WL l0 ≠ [], isPropFO(n, l, bq(t)), ¬ t0 ∈ writers(l0), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦

• Proof step 16:

  t ≠ t0, bq(t0) |WL l0 ≠ [], isPropFO(n, l, bq(t)), ¬ t0 ∈ writers(l0), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦

Rule argument

Additional information

Proof step 14 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  bq(t0) |WL l0 ≠ [], isPropFO(n, l, bq(t0)), ¬ t0 ∈ writers(l0), ∀ l, t. (t ∈ writers(l) → (∀ l0. l0 ≠ l → ¬ t ∈ writers(l0))) ∧ (¬ t ∈ writers(l) → pq(l) |W | t = [] ∧ bq(t) |WL l = []) ⊦

Rule argument

Additional information

Used simplifier rules: isPropFO-restrict-delpos: ⊦ isPropFO(n, l, bel) → bel -p n |WL l0 = bel |WL l0;

Proof step 15 from proof writersOk-prop-FO (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: l0, t0

Proof step 16 from proof writersOk-prop-FO (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: l0, t0