Proof step 1 from proof merge-owned (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  a0 ≠ a, (a[hP] * hR)[rs, f, g], (a0[hP0] * hR0)[rs, f, g], ¬ (a[hP] * a0[hP0] * true)[rs, f, g] ⊦

Rule argument

Additional information

Proof step 2 from proof merge-owned (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  a0 ≠ a, (a0[hP0] * hR0)[rs, f, g], (a[hP] * hR)[rs, f, g], ¬ (a[hP] * a0[hP0] * true)[rs, f, g] ⊦

Rule argument

Additional information

Proof step 3 from proof merge-owned (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ (hR0 * hR1)[rs, f, g] ↔ (∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ hR0[rs0, f, g] ∧ hR1[rs1, f, g])

• Proof step 5:

  a0 ≠ a, (a0[hP0] * hR0)[rs, f, g], ¬ (a[hP] * a0[hP0] * true)[rs, f, g], ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ a[hP][rs0, f, g] ∧ hR[rs1, f, g] ⊦

Rule argument

Additional information

Used terms: rs, hR, g, a[hP], f

Proof step 4 from proof merge-owned (previous step) (back to tree)

A spec-lemma

Proof step 5 from proof merge-owned (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  rs0 ∩ rs1 = ∅, a0 ≠ a, a[hP][rs0, f, g], hR[rs1, f, g], (a0[hP0] * hR0)[rs0 ∪ rs1, f, g], ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g] ⊦

Rule argument

Additional information

Proof step 6 from proof merge-owned (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ (hR0 * hR1)[rs, f, g] ↔ (∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ hR0[rs0, f, g] ∧ hR1[rs1, f, g])

• Proof step 8:

  rs0 ∩ rs1 = ∅, a0 ≠ a, hR[rs1, f, g], a[hP][rs0, f, g], ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ∃ rs2, rs3. rs2 ∪ rs3 = rs0 ∪ rs1 ∧ rs2 ∩ rs3 = ∅ ∧ a0[hP0][rs2, f, g] ∧ hR0[rs3, f, g] ⊦

Rule argument

Additional information

Used terms: rs0 ∪ rs1, hR0, g, a0[hP0], f

Proof step 7 from proof merge-owned (previous step) (back to tree)

A spec-lemma

Proof step 8 from proof merge-owned (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  rs2 ∩ rs3 = ∅, rs0 ∩ rs1 = ∅, rs2 ∪ rs3 = rs0 ∪ rs1, a0 ≠ a, a0[hP0][rs2, f, g], hR0[rs3, f, g], a[hP][rs0, f, g], hR[rs1, f, g], ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g] ⊦

Rule argument

Additional information

Proof step 9 from proof merge-owned (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 10:

  rs0 ∩ rs1 = ∅, rs2 ∩ rs3 = ∅, rs2 ∪ rs3 = rs0 ∪ rs1, a0 ≠ a, hR[rs1, f, g], a[hP][rs0, f, g], hR0[rs3, f, g], a0[hP0][rs2, f, g], rs2 ⊆ rs1, ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g] ⊦

• Proof step 19:

  rs0 ∩ rs1 = ∅, rs2 ∩ rs3 = ∅, rs2 ∪ rs3 = rs0 ∪ rs1, a0 ≠ a, hR[rs1, f, g], a[hP][rs0, f, g], hR0[rs3, f, g], a0[hP0][rs2, f, g], ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ¬ rs2 ⊆ rs1 ⊦

Rule argument

Additional information

Used formulas: rs2 ⊆ rs1

Proof step 10 from proof merge-owned (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

  ⊦ s ⊆ s0 ↔ (∃ s1. s ∪ s1 = s0 ∧ s ∩ s1 = ∅)

• Proof step 12:

  rs2 ∩ rs3 = ∅, rs0 ∩ rs1 = ∅, rs2 ∪ rs3 = rs0 ∪ rs1, a0 ≠ a, a0[hP0][rs2, f, g], hR0[rs3, f, g], a[hP][rs0, f, g], hR[rs1, f, g], ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ∃ s1. rs2 ∪ s1 = rs1 ∧ rs2 ∩ s1 = ∅ ⊦

Rule argument

Additional information

Used terms: rs1, rs2

Proof step 11 from proof merge-owned (previous step) (back to tree)

A spec-lemma

Proof step 12 from proof merge-owned (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  rs2 ∩ rs3 = ∅, rs2 ∩ rs = ∅, rs0 ∩ rs2 = ∅, rs0 ∩ rs = ∅, rs2 ∪ rs3 = rs0 ∪ rs2 ∪ rs, a0 ≠ a, hR[rs2 ∪ rs, f, g], a[hP][rs0, f, g], hR0[rs3, f, g], a0[hP0][rs2, f, g], ¬ (a[hP] * a0[hP0] * true)[rs2 ∪ rs3, f, g] ⊦

Rule argument

Additional information

Used simplifier rules: int-c: ⊦ s ∩ s0 = s0 ∩ s; from specification set-union[refset] int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; from specification set-union[refset] unc: ⊦ s ∪ s0 = s0 ∪ s; from specification set-union[refset] una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1; from specification set-union[refset] ⊦ s ∩ (s0 ∪ s1) = ∅ ↔ s ∩ s0 = ∅ ∧ s ∩ s1 = ∅; (ws)

Proof step 13 from proof merge-owned (previous step) (back to tree)

User applied rule subst equation

And obtained the following new goals

• Proof step 14:

  rs0 ∩ rs = ∅, rs0 ∩ rs2 = ∅, rs2 ∩ rs = ∅, rs2 ∩ rs3 = ∅, a0 ≠ a, a0[hP0][rs2, f, g], hR0[rs3, f, g], a[hP][rs0, f, g], hR[rs2 ∪ rs, f, g], ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs2 ∪ rs, f, g] ⊦

Rule argument

Additional information

Proof step 14 from proof merge-owned (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 15:

  ⊦ hR0[rs0, f, g] ∧ hR1[rs1, f, g] ∧ rs0 ∩ rs1 = ∅ → (hR0 * hR1)[rs0 ∪ rs1, f, g]

• Proof step 16:

  rs2 ∩ rs3 = ∅, rs2 ∩ rs = ∅, rs0 ∩ rs2 = ∅, rs0 ∩ rs = ∅, a0 ≠ a, hR[rs2 ∪ rs, f, g], a[hP][rs0, f, g], hR0[rs3, f, g], a0[hP0][rs2, f, g], ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs2 ∪ rs, f, g] ⊦ a[hP][rs0, f, g] ∧ (a0[hP0] * true)[rs2 ∪ rs, f, g] ∧ rs0 ∩ (rs2 ∪ rs) = ∅

Rule argument

Additional information

Used terms: rs0, a0[hP0] * true, g, a[hP], f, rs2 ∪ rs

Proof step 15 from proof merge-owned (previous step) (back to tree)

A spec-lemma

Proof step 16 from proof merge-owned (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

  rs0 ∩ rs = ∅, rs0 ∩ rs2 = ∅, rs2 ∩ rs = ∅, rs2 ∩ rs3 = ∅, a0 ≠ a, a0[hP0][rs2, f, g], hR0[rs3, f, g], a[hP][rs0, f, g], hR[rs2 ∪ rs, f, g], ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs2 ∪ rs, f, g], ¬ (a0[hP0] * true)[rs2 ∪ rs, f, g] ⊦

Rule argument

Additional information

Used simplifier rules: int-c: ⊦ s ∩ s0 = s0 ∩ s; from specification set-union[refset] int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; from specification set-union[refset] unc: ⊦ s ∪ s0 = s0 ∪ s; from specification set-union[refset] una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1; from specification set-union[refset] ⊦ s ∩ (s0 ∪ s1) = ∅ ↔ s ∩ s0 = ∅ ∧ s ∩ s1 = ∅; (ws)

Proof step 17 from proof merge-owned (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 18:

  ⊦ hR0[rs0, f, g] ∧ hR1[rs1, f, g] ∧ rs0 ∩ rs1 = ∅ → (hR0 * hR1)[rs0 ∪ rs1, f, g]

Rule argument

Additional information

Used simplifier rules: ⊦ true[rs, f, g]; (s) Used terms: rs2, true, g, a0[hP0], f, rs

Proof step 18 from proof merge-owned (previous step) (back to tree)

A spec-lemma

Proof step 19 from proof merge-owned (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 20:

  ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2)

• Proof step 21:

  rs2 ∩ rs3 = ∅, rs0 ∩ rs1 = ∅, rs2 ∪ rs3 = rs0 ∪ rs1, a0 ≠ a, a0[hP0][rs2, f, g], hR0[rs3, f, g], a[hP][rs0, f, g], hR[rs1, f, g], ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ∃ r. ¬(r ∈ rs2 → r ∈ rs1) ⊦

Rule argument

Additional information

Used terms: rs1, rs2

Proof step 20 from proof merge-owned (previous step) (back to tree)

A spec-lemma

Proof step 21 from proof merge-owned (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  rs0 ∩ rs1 = ∅, rs2 ∩ rs3 = ∅, rs2 ∪ rs3 = rs0 ∪ rs1, a0 ≠ a, hR[rs1, f, g], a[hP][rs0, f, g], hR0[rs3, f, g], a0[hP0][rs2, f, g], r ∈ rs2, ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ¬ r ∈ rs1 ⊦

Rule argument

Additional information

Proof step 22 from proof merge-owned (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 23:

  ⊦ hR[rs, f, g] ↔ hR(mk(rs, f, g))

• Proof step 24:

  rs2 ∩ rs3 = ∅, rs0 ∩ rs1 = ∅, rs2 ∪ rs3 = rs0 ∪ rs1, a0 ≠ a, a0[hP0][rs2, f, g], hR0[rs3, f, g], a[hP](mk(rs0, f, g)), hR[rs1, f, g], r ∈ rs2, ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ¬ r ∈ rs1 ⊦

Rule argument

Additional information

Used terms: g, f, rs0, a[hP]

Proof step 23 from proof merge-owned (previous step) (back to tree)

A spec-lemma

Proof step 24 from proof merge-owned (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 25:

  ⊦ hR[rs, f, g] ↔ hR(mk(rs, f, g))

• Proof step 26:

  rs0 ∩ rs1 = ∅, rs2 ∩ rs3 = ∅, rs2 ∪ rs3 = rs0 ∪ rs1, a0 ≠ a, hR[rs1, f, g], a[hP](mk(rs0, f, g)), hR0[rs3, f, g], a0[hP0](mk(rs2, f, g)), r ∈ rs2, ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ¬ r ∈ rs1 ⊦

Rule argument

Additional information

Used terms: g, f, rs2, a0[hP0]

Proof step 25 from proof merge-owned (previous step) (back to tree)

A spec-lemma

Proof step 26 from proof merge-owned (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 27:

  ⊦ a[hP](mk(rs, f, g)) ↔ hP(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a)

• Proof step 28:

  rs2 ∩ rs3 = ∅, rs0 ∩ rs1 = ∅, rs2 ∪ rs3 = rs0 ∪ rs1, a0 ≠ a, hR0[rs3, f, g], a[hP](mk(rs0, f, g)), hR[rs1, f, g], hP0(mk(rs2, f)), r ∈ rs2, ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ¬ r ∈ rs1, ∀ r. r ∈ rs2 → g(r) = a0 ⊦

Rule argument

Additional information

Used terms: a0, hP0, rs2, g, f

Proof step 27 from proof merge-owned (previous step) (back to tree)

Lemma lift-aux

Proof step 28 from proof merge-owned (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  rs0 ∩ rs1 = ∅, rs2 ∩ rs3 = ∅, rs2 ∪ rs3 = rs0 ∪ rs1, a0 ≠ a, hR[rs1, f, g], a[hP](mk(rs0, f, g)), hR0[rs3, f, g], hP0(mk(rs2, f)), r ∈ rs2, ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ¬ r ∈ rs1, ∀ r. r ∈ rs2 → g(r) = a0 ⊦

Rule argument

Additional information

Proof step 29 from proof merge-owned (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 30:

  ⊦ a[hP](mk(rs, f, g)) ↔ hP(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a)

• Proof step 31:

  rs2 ∩ rs3 = ∅, rs0 ∩ rs1 = ∅, rs2 ∪ rs3 = rs0 ∪ rs1, a0 ≠ a, hR0[rs3, f, g], hR[rs1, f, g], hP(mk(rs0, f)), hP0(mk(rs2, f)), r ∈ rs2, ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ¬ r ∈ rs1, ∀ r. r ∈ rs2 → g(r) = a0, ∀ r. r ∈ rs0 → g(r) = a ⊦

Rule argument

Additional information

Used terms: a, hP, rs0, g, f

Proof step 30 from proof merge-owned (previous step) (back to tree)

Lemma lift-aux

Proof step 31 from proof merge-owned (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

  rs0 ∩ rs1 = ∅, rs2 ∩ rs3 = ∅, rs2 ∪ rs3 = rs0 ∪ rs1, a0 ≠ a, hR[rs1, f, g], hR0[rs3, f, g], hP(mk(rs0, f)), hP0(mk(rs2, f)), r ∈ rs2, ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ¬ r ∈ rs1, ∀ r. r ∈ rs0 → g(r) = a, ∀ r. r ∈ rs2 → g(r) = a0 ⊦

Rule argument

Additional information

Proof step 32 from proof merge-owned (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 33:

  ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)

• Proof step 34:

  rs2 ∩ rs3 = ∅, rs0 ∩ rs1 = ∅, a0 ≠ a, hR0[rs3, f, g], hR[rs1, f, g], hP(mk(rs0, f)), hP0(mk(rs2, f)), r ∈ rs2, ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ¬ r ∈ rs1, ∀ r. r ∈ rs2 → g(r) = a0, ∀ r. r ∈ rs0 → g(r) = a, ∀ r. r ∈ rs2 ∪ rs3 ↔ r ∈ rs0 ∪ rs1 ⊦

Rule argument

Additional information

Used terms: rs0 ∪ rs1, rs2 ∪ rs3

Proof step 33 from proof merge-owned (previous step) (back to tree)

A spec-lemma

Proof step 34 from proof merge-owned (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 35:

  rs0 ∩ rs1 = ∅, rs2 ∩ rs3 = ∅, a0 ≠ a, hR[rs1, f, g], hR0[rs3, f, g], hP(mk(rs0, f)), hP0(mk(rs2, f)), r ∈ rs2, ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ¬ r ∈ rs1, r ∈ rs0 → g(r) = a, ∀ r. r ∈ rs2 ∪ rs3 ↔ r ∈ rs0 ∪ rs1, ∀ r. r ∈ rs0 → g(r) = a, ∀ r. r ∈ rs2 → g(r) = a0 ⊦

Rule argument

Additional information

Used terms: r

Proof step 35 from proof merge-owned (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 36:

  rs2 ∩ rs3 = ∅, rs0 ∩ rs1 = ∅, a0 ≠ a, hR0[rs3, f, g], hR[rs1, f, g], hP(mk(rs0, f)), hP0(mk(rs2, f)), r ∈ rs2, ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs1, f, g], ¬ r ∈ rs1, r ∈ rs0 → g(r) = a, r ∈ rs2 → g(r) = a0, ∀ r. r ∈ rs2 → g(r) = a0, ∀ r. r ∈ rs0 → g(r) = a, ∀ r. r ∈ rs2 ∪ rs3 ↔ r ∈ rs0 ∪ rs1 ⊦

Rule argument

Additional information

Used terms: r

Proof step 36 from proof merge-owned (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  rs2 ∩ rs3 = ∅, rs0 ∩ rs1 = ∅, g(r) ≠ a, hR0[rs3, f, g], hR[rs1, f, g], hP(mk(rs0, f)), hP0(mk(rs2, f)), r ∈ rs2, ¬ (a[hP] * g(r)[hP0] * true)[rs0 ∪ rs1, f, g], ¬ r ∈ rs0, ¬ r ∈ rs1, ∀ r0. r0 ∈ rs2 → g(r0) = g(r), ∀ r. r ∈ rs0 → g(r) = a, ∀ r. r ∈ rs2 ∪ rs3 ↔ r ∈ rs0 ∪ rs1 ⊦

Rule argument

Additional information

Proof step 37 from proof merge-owned (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ s → (a ∈ s ∪ s0 ↔ a ∈ s0); (s) ⊦ a ∈ s ∪ s0 ↔ ¬(¬ a ∈ s ∧ ¬ a ∈ s0); (ss) 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