Proof step 1 from proof add-subset (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬((h0 ++ r)(r := d) ⊆ h ↔ h0 ⊆ h ∧ r ∈ h ∧ h(r) = d), r ≠ null, ¬ r ∈ h0 ⊦

Rule argument

Additional information

Proof step 2 from proof add-subset (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 3:

  ⊦ h0 ⊆ h1 ↔ (∀ r. (r ∈ h0 → r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r)))

• Proof step 4:

  r ≠ null, ¬ r ∈ h0 ⊦ (∀ r0. (r0 ∈ (h0 ++ r)(r := d) → r0 ∈ h) ∧ (r0 ∈ (h0 ++ r)(r := d) → (h0 ++ r)(r := d)(r0) = h(r0))) ↔ h0 ⊆ h ∧ r ∈ h ∧ h(r) = d

Rule argument

Additional information

Used terms: (h0 ++ r)(r := d), h

Proof step 3 from proof add-subset (previous step) (back to tree)

Lemma subset

Proof step 4 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  r ≠ null, ¬ r ∈ h0 ⊦ (∀ r0. (r0 = r ∨ r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ d;(h0 ++ r)(r0)) = h(r0))) ↔ h0 ⊆ h ∧ r ∈ h ∧ h(r) = d

Rule argument

Additional information

Used simplifier rules: r0 ≠ null, ¬ r0 ∈ h ⊦ r ∈ h ++ r0 ↔ r = r0 ∨ r ∈ h; (s) ⊦ h(r0 := o)(r) = (r = r0 ∧ r0 ∈ h ⊃ o;h(r)); (s) ⊦ r ∈ h(r0 := o) ↔ r ∈ h; (s)

Proof step 5 from proof add-subset (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

  ⊦ h0 ⊆ h1 ↔ (∀ r. (r ∈ h0 → r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r)))

• Proof step 7:

  r ≠ null, ¬ r ∈ h0 ⊦ (∀ r0. (r0 = r ∨ r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ d;(h0 ++ r)(r0)) = h(r0))) ↔ (∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r))) ∧ r ∈ h ∧ h(r) = d

Rule argument

Additional information

Used terms: h0, h

Proof step 6 from proof add-subset (previous step) (back to tree)

Lemma subset

Proof step 7 from proof add-subset (previous step) (back to tree)

Heuristic cut applied rule cut formula

And obtained the following new goals

• Proof step 8:

  r ≠ null, r ∈ h, ¬ r ∈ h0 ⊦ (∀ r0. (r0 = r ∨ r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ d;(h0 ++ r)(r0)) = h(r0))) ↔ (∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r))) ∧ r ∈ h ∧ h(r) = d

• Proof step 32:

  r ≠ null, ¬ r ∈ h, ¬ r ∈ h0 ⊦ (∀ r0. (r0 = r ∨ r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ d;(h0 ++ r)(r0)) = h(r0))) ↔ (∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r))) ∧ r ∈ h ∧ h(r) = d

Rule argument

Additional information

Used formulas: r ∈ h

Proof step 8 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  r ∈ h, ¬ r ∈ h0 ⊦ (∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ d;(h0 ++ r)(r0)) = h(r0))) ↔ (∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r))) ∧ h(r) = d

Rule argument

Additional information

Used simplifier rules: ⊦ r ∈ h → r ≠ null; (s)

Proof step 9 from proof add-subset (previous step) (back to tree)

Heuristic cut applied rule cut formula

And obtained the following new goals

• Proof step 10:

  h(r) = d, r ∈ h, ¬ r ∈ h0 ⊦ (∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ d;(h0 ++ r)(r0)) = h(r0))) ↔ (∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r))) ∧ h(r) = d

• Proof step 30:

  h(r) ≠ d, r ∈ h, ¬ r ∈ h0 ⊦ (∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ d;(h0 ++ r)(r0)) = h(r0))) ↔ (∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r))) ∧ h(r) = d

Rule argument

Additional information

Used formulas: h(r) = d

Proof step 10 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  r ∈ h, ¬ r ∈ h0 ⊦ (∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ h(r);(h0 ++ r)(r0)) = h(r0))) ↔ (∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r)))

Rule argument

Additional information

Proof step 11 from proof add-subset (previous step) (back to tree)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

• Proof step 12:

  r ∈ h, ¬ r ∈ h0, ∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ h(r);(h0 ++ r)(r0)) = h(r0)) ⊦ ∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r))

• Proof step 20:

  r ∈ h, ¬ r ∈ h0, ∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r)) ⊦ ∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ h(r);(h0 ++ r)(r0)) = h(r0))

Rule argument

Additional information

Proof step 12 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  r ∈ h, ¬ r ∈ h0, ∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ h(r);(h0 ++ r)(r0)) = h(r0)) ⊦ (r0 ∈ h0 → r0 ∈ h) ∧ (r0 ∈ h0 → h0(r0) = h(r0))

Rule argument

Additional information

Proof step 13 from proof add-subset (previous step) (back to tree)

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

• Proof step 14:

  r ∈ h, r0 ∈ h0, ¬ r ∈ h0, ∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ h(r);(h0 ++ r)(r0)) = h(r0)) ⊦ (r0 ∈ h0 → r0 ∈ h) ∧ (r0 ∈ h0 → h0(r0) = h(r0))

Rule argument

Additional information

Used formulas: r0 ∈ h0

Proof step 14 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  r ∈ h, r0 ∈ h0, ¬ r ∈ h0, ∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ h(r);(h0 ++ r)(r0)) = h(r0)) ⊦ r0 ∈ h ∧ h0(r0) = h(r0)

Rule argument

Additional information

Proof step 15 from proof add-subset (previous step) (back to tree)

Heuristic cut applied rule cut formula

And obtained the following new goals

• Proof step 16:

  r ∈ h, r0 ∈ h0, r0 ∈ h, ¬ r ∈ h0, ∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ h(r);(h0 ++ r)(r0)) = h(r0)) ⊦ r0 ∈ h ∧ h0(r0) = h(r0)

• Proof step 18:

  r ∈ h, r0 ∈ h0, ¬ r ∈ h0, ¬ r0 ∈ h, ∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ h(r);(h0 ++ r)(r0)) = h(r0)) ⊦ r0 ∈ h ∧ h0(r0) = h(r0)

Rule argument

Additional information

Used formulas: r0 ∈ h

Proof step 16 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

  h0(r0) ≠ h(r0), r ∈ h, r0 ∈ h, r0 ∈ h0, ¬ r ∈ h0, ∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ h(r);(h0 ++ r)(r0)) = h(r0)) ⊦

Rule argument

Additional information

Proof step 17 from proof add-subset (previous step) (back to tree)

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ r ∈ h → r ≠ null; (s) r ∈ h, r0 ≠ null, ¬ r0 ∈ h ⊦ (h ++ r0)(r) = h(r); (s) Used terms: r0

Proof step 18 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  r ∈ h, r0 ∈ h0, ¬ r ∈ h0, ¬ r0 ∈ h, ∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ h(r);(h0 ++ r)(r0)) = h(r0)) ⊦

Rule argument

Additional information

Proof step 19 from proof add-subset (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ r ∈ h → r ≠ null; (s) r ∈ h, r0 ≠ null, ¬ r0 ∈ h ⊦ (h ++ r0)(r) = h(r); (s) Used terms: r0

Proof step 20 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  r ∈ h, ¬ r ∈ h0, ∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r)) ⊦ (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ h(r);(h0 ++ r)(r0)) = h(r0))

Rule argument

Additional information

Proof step 21 from proof add-subset (previous step) (back to tree)

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

• Proof step 22:

  r0 ≠ r, r ∈ h, ¬ r ∈ h0, ∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r)) ⊦ (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ h(r);(h0 ++ r)(r0)) = h(r0))

Rule argument

Additional information

Used formulas: r0 = r

Proof step 22 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  r0 ≠ r, r ∈ h, ¬ r ∈ h0, ∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r)) ⊦ (r0 ∈ h0 → r0 ∈ h) ∧ (r0 ∈ h0 → h0(r0) = h(r0))

Rule argument

Additional information

Used simplifier rules: r ∈ h, r0 ≠ null, ¬ r0 ∈ h ⊦ (h ++ r0)(r) = h(r); (s) ⊦ r ∈ h → r ≠ null; (s)

Proof step 23 from proof add-subset (previous step) (back to tree)

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

• Proof step 24:

  r0 ≠ r, r ∈ h, r0 ∈ h0, ¬ r ∈ h0, ∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r)) ⊦ (r0 ∈ h0 → r0 ∈ h) ∧ (r0 ∈ h0 → h0(r0) = h(r0))

Rule argument

Additional information

Used formulas: r0 ∈ h0

Proof step 24 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  r0 ≠ r, r ∈ h, r0 ∈ h0, ¬ r ∈ h0, ∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r)) ⊦ r0 ∈ h ∧ h0(r0) = h(r0)

Rule argument

Additional information

Proof step 25 from proof add-subset (previous step) (back to tree)

Heuristic cut applied rule cut formula

And obtained the following new goals

• Proof step 26:

  r0 ≠ r, r ∈ h, r0 ∈ h0, r0 ∈ h, ¬ r ∈ h0, ∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r)) ⊦ r0 ∈ h ∧ h0(r0) = h(r0)

• Proof step 28:

  r0 ≠ r, r ∈ h, r0 ∈ h0, ¬ r ∈ h0, ¬ r0 ∈ h, ∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r)) ⊦ r0 ∈ h ∧ h0(r0) = h(r0)

Rule argument

Additional information

Used formulas: r0 ∈ h

Proof step 26 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  h0(r0) ≠ h(r0), r0 ≠ r, r ∈ h, r0 ∈ h, r0 ∈ h0, ¬ r ∈ h0, ∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r)) ⊦

Rule argument

Additional information

Proof step 27 from proof add-subset (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: r0

Proof step 28 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  r0 ≠ r, r ∈ h, r0 ∈ h0, ¬ r ∈ h0, ¬ r0 ∈ h, ∀ r. (r ∈ h0 → r ∈ h) ∧ (r ∈ h0 → h0(r) = h(r)) ⊦

Rule argument

Additional information

Proof step 29 from proof add-subset (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: r0

Proof step 30 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  h(r) ≠ d, r ∈ h, ¬ r ∈ h0, ∀ r0. (r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ d;(h0 ++ r)(r0)) = h(r0)) ⊦

Rule argument

Additional information

Proof step 31 from proof add-subset (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: r

Proof step 32 from proof add-subset (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  r ≠ null, ¬ r ∈ h, ¬ r ∈ h0, ∀ r0. (r0 = r ∨ r0 ∈ h0 → r0 ∈ h) ∧ (r0 = r ∨ r0 ∈ h0 → (r0 = r ⊃ d;(h0 ++ r)(r0)) = h(r0)) ⊦

Rule argument

Additional information

Proof step 33 from proof add-subset (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: r