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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 2:

  ⊦ ((hP0 * hP1))(mk(rs, f)) ↔ (∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ hP0(mk(rs0, f)) ∧ hP1(mk(rs1, f)))

Rule argument

Additional information

Used simplifier rules: apply: ⊦ hP[rs, f] ↔ hP(mk(rs, f));

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

User applied rule case distinction

And obtained the following new goals

• Proof step 3:

  ((hP0 * hP1))(mk(rs, f)) ⊦ ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ hP0(mk(rs0, f)) ∧ hP1(mk(rs1, f))

• Proof step 38:

  ¬ ((hP0 * hP1))(mk(rs, f)), ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ hP0(mk(rs0, f)) ∧ hP1(mk(rs1, f)) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ ((hP0 * hP1))(h) ↔ (∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1))

• Proof step 5:

  ∃ h0, h1. h0 ⊥ h1 ∧ mk(rs, f) = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1) ⊦ ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ hP0(mk(rs0, f)) ∧ hP1(mk(rs1, f))

Rule argument

Additional information

Used terms: mk(rs, f), hP1, hP0

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  mk(rs, f) = h0 ∪ h1, hP0(h0), hP1(h1), h0 ⊥ h1 ⊦ ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ hP0(mk(rs0, f)) ∧ hP1(mk(rs1, f))

Rule argument

Additional information

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 7:

  ⊦ ∃ rs, f. h = mk(rs, f)

• Proof step 8:

  mk(rs, f) = h0 ∪ h1, hP0(h0), hP1(h1), h0 ⊥ h1, ∃ rs, f. h0 = mk(rs, f) ⊦ ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ hP0(mk(rs0, f)) ∧ hP1(mk(rs1, f))

Rule argument

Additional information

Used terms: h0

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

Lemma mk-exists

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  mk(rs, f) = mk(rs0, f0) ∪ h1, hP0(mk(rs0, f0)), hP1(h1), mk(rs0, f0) ⊥ h1 ⊦ ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ hP0(mk(rs0, f)) ∧ hP1(mk(rs1, f))

Rule argument

Additional information

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 10:

  ⊦ ∃ rs, f. h = mk(rs, f)

• Proof step 11:

  mk(rs, f) = mk(rs0, f0) ∪ h1, hP0(mk(rs0, f0)), hP1(h1), mk(rs0, f0) ⊥ h1, ∃ rs, f. h1 = mk(rs, f) ⊦ ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ hP0(mk(rs0, f)) ∧ hP1(mk(rs1, f))

Rule argument

Additional information

Used terms: h1

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

Lemma mk-exists

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, hP0(mk(rs0, f0)), hP1(mk(rs1, f1)) ⊦ ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ hP0(mk(rs0, f)) ∧ hP1(mk(rs1, f))

Rule argument

Additional information

Used simplifier rules: disj: ⊦ mk(rs0, f0) ⊥ mk(rs1, f1) ↔ rs0 ∩ rs1 = ∅;

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

User applied rule exists right

And obtained the following new goals

• Proof step 13:

  mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, hP0(mk(rs0, f0)), hP1(mk(rs1, f1)) ⊦ rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ hP0(mk(rs0, f)) ∧ hP1(mk(rs1, f))

Rule argument

Additional information

Used terms: rs0, rs1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, hP0(mk(rs0, f0)), hP1(mk(rs1, f1)) ⊦ rs0 ∪ rs1 = rs ∧ hP0(mk(rs0, f)) ∧ hP1(mk(rs1, f))

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 15:

  mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, rs0 ∪ rs1 ≠ rs, hP0(mk(rs0, f0)), hP1(mk(rs1, f1)) ⊦

• Proof step 22:

  mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), ¬ hP0(mk(rs0, f)) ⊦

• Proof step 30:

  mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), ¬ hP1(mk(rs1, f)) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

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

• Proof step 17:

  rs0 ∩ rs1 = ∅, rs0 ∪ rs1 ≠ rs, hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), ∀ r. (r ∈ mk(rs, f) ↔ r ∈ mk(rs0, f0) ∪ mk(rs1, f1)) ∧ (r ∈ mk(rs, f) → mk(rs, f)[r] = (mk(rs0, f0) ∪ mk(rs1, f1))[r]) ⊦

Rule argument

Additional information

Used terms: mk(rs0, f0) ∪ mk(rs1, f1), mk(rs, f)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  rs0 ∩ rs1 = ∅, rs0 ∪ rs1 ≠ rs, hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), ∀ r. (r ∈ rs ↔ r ∈ rs0 ∨ r ∈ rs1) ∧ (r ∈ rs → f(r) = (mk(rs0, f0) ∪ mk(rs1, f1))[r]) ⊦

Rule argument

Additional information

Used simplifier rules: in-uni: ⊦ rs0 ∩ rs1 = ∅ → (r ∈ mk(rs0, f0) ∪ mk(rs1, f1) ↔ r ∈ rs0 ∨ r ∈ rs1); in-mk: ⊦ r ∈ mk(rs, f) ↔ r ∈ rs; at-mk: ⊦ r ∈ rs → mk(rs, f)[r] = f(r); int-c: ⊦ s ∩ s0 = s0 ∩ s; from specification set-union[refset] int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; from specification set-union[refset]

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

  ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)

• Proof step 20:

  rs0 ∩ rs1 = ∅, hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), ∀ r. (r ∈ rs ↔ r ∈ rs0 ∨ r ∈ rs1) ∧ (r ∈ rs → f(r) = (mk(rs0, f0) ∪ mk(rs1, f1))[r]), ∃ a. ¬(a ∈ rs0 ∪ rs1 ↔ a ∈ rs) ⊦

Rule argument

Additional information

Used terms: rs, rs0 ∪ rs1

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  rs0 ∩ rs1 = ∅, ¬(r ∈ rs0 ∪ rs1 ↔ r ∈ rs), hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), ∀ r. (r ∈ rs ↔ r ∈ rs0 ∨ r ∈ rs1) ∧ (r ∈ rs → f(r) = (mk(rs0, f0) ∪ mk(rs1, f1))[r]) ⊦

Rule argument

Additional information

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ s ∪ s0 ↔ ¬(¬ a ∈ s ∧ ¬ a ∈ s0); (ss) ⊦ ¬ a ∈ s → (a ∈ s ∪ s0 ↔ a ∈ s0); (s) una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1; from specification set-union[refset] unc: ⊦ s ∪ s0 = s0 ∪ s; from specification set-union[refset] Used terms: r

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

User applied rule cut formula

And obtained the following new goals

• Proof step 23:

  mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, mk(rs0, f) ≠ mk(rs0, f0), hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), ¬ hP0(mk(rs0, f)) ⊦

Rule argument

Additional information

Used formulas: mk(rs0, f) = mk(rs0, f0)

Proof step 23 from proof sep (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 24:

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

• Proof step 25:

  mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), ¬ hP0(mk(rs0, f)), ∃ r. ¬((r ∈ mk(rs0, f) ↔ r ∈ mk(rs0, f0)) ∧ (r ∈ mk(rs0, f) → mk(rs0, f)[r] = mk(rs0, f0)[r])) ⊦

Rule argument

Additional information

Used terms: mk(rs0, f0), mk(rs0, f)

Proof step 24 from proof sep (previous step) (back to tree)

A spec-lemma

Proof step 25 from proof sep (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, f(r) ≠ f0(r), hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), r ∈ rs0, ¬ hP0(mk(rs0, f)) ⊦

Rule argument

Additional information

Used simplifier rules: in-mk: ⊦ r ∈ mk(rs, f) ↔ r ∈ rs; at-mk: ⊦ r ∈ rs → mk(rs, f)[r] = f(r);

Proof step 26 from proof sep (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 27:

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

• Proof step 28:

  rs0 ∩ rs1 = ∅, f(r) ≠ f0(r), hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), r ∈ rs0, ¬ hP0(mk(rs0, f)), ∀ r. (r ∈ mk(rs, f) ↔ r ∈ mk(rs0, f0) ∪ mk(rs1, f1)) ∧ (r ∈ mk(rs, f) → mk(rs, f)[r] = (mk(rs0, f0) ∪ mk(rs1, f1))[r]) ⊦

Rule argument

Additional information

Used terms: mk(rs0, f0) ∪ mk(rs1, f1), mk(rs, f)

Proof step 27 from proof sep (previous step) (back to tree)

A spec-lemma

Proof step 28 from proof sep (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  rs0 ∩ rs1 = ∅, f(r) ≠ f0(r), hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), r ∈ rs0, ¬ hP0(mk(rs0, f)), ∀ r0. (r0 ∈ rs ↔ r0 ∈ rs0 ∨ r0 ∈ rs1) ∧ (r0 ∈ rs → f(r0) = (mk(rs0, f0) ∪ mk(rs1, f1))[r0]) ⊦

Rule argument

Additional information

Used simplifier rules: in-uni: ⊦ rs0 ∩ rs1 = ∅ → (r ∈ mk(rs0, f0) ∪ mk(rs1, f1) ↔ r ∈ rs0 ∨ r ∈ rs1); in-mk: ⊦ r ∈ mk(rs, f) ↔ r ∈ rs; at-mk: ⊦ r ∈ rs → mk(rs, f)[r] = f(r); int-c: ⊦ s ∩ s0 = s0 ∩ s; from specification set-union[refset] int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; from specification set-union[refset]

Proof step 29 from proof sep (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: at-unil: ⊦ rs0 ∩ rs1 = ∅ ∧ r ∈ rs0 → (mk(rs0, f0) ∪ mk(rs1, f1))[r] = f0(r); int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; from specification set-union[refset] int-c: ⊦ s ∩ s0 = s0 ∩ s; from specification set-union[refset] Used terms: r

Proof step 30 from proof sep (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 31:

  mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, mk(rs1, f) ≠ mk(rs1, f1), hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), ¬ hP1(mk(rs1, f)) ⊦

Rule argument

Additional information

Used formulas: mk(rs1, f) = mk(rs1, f1)

Proof step 31 from proof sep (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 32:

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

• Proof step 33:

  mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), ¬ hP1(mk(rs1, f)), ∃ r. ¬((r ∈ mk(rs1, f) ↔ r ∈ mk(rs1, f1)) ∧ (r ∈ mk(rs1, f) → mk(rs1, f)[r] = mk(rs1, f1)[r])) ⊦

Rule argument

Additional information

Used terms: mk(rs1, f1), mk(rs1, f)

Proof step 32 from proof sep (previous step) (back to tree)

A spec-lemma

Proof step 33 from proof sep (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, f(r) ≠ f1(r), hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), r ∈ rs1, ¬ hP1(mk(rs1, f)) ⊦

Rule argument

Additional information

Used simplifier rules: in-mk: ⊦ r ∈ mk(rs, f) ↔ r ∈ rs; at-mk: ⊦ r ∈ rs → mk(rs, f)[r] = f(r);

Proof step 34 from proof sep (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 35:

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

• Proof step 36:

  rs0 ∩ rs1 = ∅, f(r) ≠ f1(r), hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), r ∈ rs1, ¬ hP1(mk(rs1, f)), ∀ r. (r ∈ mk(rs, f) ↔ r ∈ mk(rs0, f0) ∪ mk(rs1, f1)) ∧ (r ∈ mk(rs, f) → mk(rs, f)[r] = (mk(rs0, f0) ∪ mk(rs1, f1))[r]) ⊦

Rule argument

Additional information

Used terms: mk(rs0, f0) ∪ mk(rs1, f1), mk(rs, f)

Proof step 35 from proof sep (previous step) (back to tree)

A spec-lemma

Proof step 36 from proof sep (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  rs0 ∩ rs1 = ∅, f(r) ≠ f1(r), hP0(mk(rs0, f0)), hP1(mk(rs1, f1)), r ∈ rs1, ¬ hP1(mk(rs1, f)), ∀ r0. (r0 ∈ rs ↔ r0 ∈ rs0 ∨ r0 ∈ rs1) ∧ (r0 ∈ rs → f(r0) = (mk(rs0, f0) ∪ mk(rs1, f1))[r0]) ⊦

Rule argument

Additional information

Used simplifier rules: in-uni: ⊦ rs0 ∩ rs1 = ∅ → (r ∈ mk(rs0, f0) ∪ mk(rs1, f1) ↔ r ∈ rs0 ∨ r ∈ rs1); in-mk: ⊦ r ∈ mk(rs, f) ↔ r ∈ rs; at-mk: ⊦ r ∈ rs → mk(rs, f)[r] = f(r); int-c: ⊦ s ∩ s0 = s0 ∩ s; from specification set-union[refset] int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; from specification set-union[refset]

Proof step 37 from proof sep (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: at-unir: ⊦ rs0 ∩ rs1 = ∅ ∧ r ∈ rs1 → (mk(rs0, f0) ∪ mk(rs1, f1))[r] = f1(r); int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; from specification set-union[refset] int-c: ⊦ s ∩ s0 = s0 ∩ s; from specification set-union[refset] ⊦ s ∩ s0 = ∅ ∧ a ∈ s → (a ∈ s0 ↔ false); (as) Used terms: r

Proof step 38 from proof sep (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

  rs0 ∩ rs1 = ∅, hP0(mk(rs0, f)), hP1(mk(rs1, f)), ¬ ((hP0 * hP1))(mk(rs0, f) ∪ mk(rs1, f)) ⊦

Rule argument

Additional information

Used simplifier rules: uni: ⊦ rs0 ∩ rs1 = ∅ → mk(rs0 ∪ rs1, f) = mk(rs0, f) ∪ mk(rs1, f); unc: ⊦ s ∪ s0 = s0 ∪ s; from specification set-union[refset] una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1; from specification set-union[refset] int-c: ⊦ s ∩ s0 = s0 ∩ s; from specification set-union[refset] int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; from specification set-union[refset]

Proof step 39 from proof sep (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 40:

  ⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → ((hP0 * hP1))(h0 ∪ h1)

Rule argument

Additional information

Used simplifier rules: disj: ⊦ mk(rs0, f0) ⊥ mk(rs1, f1) ↔ rs0 ∩ rs1 = ∅; Used terms: mk(rs1, f), mk(rs0, f), hP1, hP0

Proof step 40 from proof sep (previous step) (back to tree)

A spec-lemma