Proof step 1 from proof mod-other-hp2 (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

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

Rule argument

Additional information

Proof step 2 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 3:

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

• Proof step 4:

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

Rule argument

Additional information

Used terms: rs, a0[hP0] * hR, g, a[hP], f

Proof step 3 from proof mod-other-hp2 (previous step) (back to tree)

A spec-lemma

Proof step 4 from proof mod-other-hp2 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

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

Rule argument

Additional information

Proof step 5 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

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

• Proof step 7:

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

Rule argument

Additional information

Used terms: rs1, hR, g, a0[hP0], f

Proof step 6 from proof mod-other-hp2 (previous step) (back to tree)

A spec-lemma

Proof step 7 from proof mod-other-hp2 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  rs ∩ rs2 = ∅, rs0 ∩ rs = ∅, rs0 ∩ rs2 = ∅, g(r) ≠ a0, g(r) ≠ a, a0[hP0][rs, f, g], hR[rs2, f, g], a[hP][rs0, f, g], ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs ∪ rs2, f(r; o), 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 8 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

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

• Proof step 10:

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

Rule argument

Additional information

Used terms: rs0, a0[hP0] * true, g, a[hP], f(r; o), rs ∪ rs2

Proof step 9 from proof mod-other-hp2 (previous step) (back to tree)

A spec-lemma

Proof step 10 from proof mod-other-hp2 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  rs ∩ rs2 = ∅, rs0 ∩ rs = ∅, rs0 ∩ rs2 = ∅, g(r) ≠ a0, g(r) ≠ a, a0[hP0][rs, f, g], hR[rs2, f, g], a[hP][rs0, f, g], ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs ∪ rs2, f(r; o), g] ⊦ a[hP][rs0, f(r; o), g] ∧ (a0[hP0] * true)[rs ∪ rs2, f(r; o), 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 11 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 12:

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

• Proof step 29:

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

Rule argument

Additional information

Proof step 12 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

  ⊦ hR[rs, f, g] ↔ hR(mk(rs, f, g))

• Proof step 14:

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

Rule argument

Additional information

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

Proof step 13 from proof mod-other-hp2 (previous step) (back to tree)

A spec-lemma

Proof step 14 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 15:

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

• Proof step 16:

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

Rule argument

Additional information

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

Proof step 15 from proof mod-other-hp2 (previous step) (back to tree)

Lemma lift-aux

Proof step 16 from proof mod-other-hp2 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

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

Rule argument

Additional information

Proof step 17 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 18:

  ⊦ hR[rs, f, g] ↔ hR(mk(rs, f, g))

• Proof step 19:

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

Rule argument

Additional information

Used terms: g, f(r; o), rs0, a[hP]

Proof step 18 from proof mod-other-hp2 (previous step) (back to tree)

A spec-lemma

Proof step 19 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 20:

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

• Proof step 21:

  rs ∩ rs2 = ∅, rs0 ∩ rs = ∅, rs0 ∩ rs2 = ∅, g(r) ≠ a0, g(r) ≠ a, a0[hP0][rs, f, g], hR[rs2, f, g], hP(mk(rs0, f)), ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs ∪ rs2, f(r; o), g], ∀ r. r ∈ rs0 → g(r) = a, ¬ hP(mk(rs0, f(r; o))) ∨ (∃ r. ¬(r ∈ rs0 → g(r) = a)) ⊦

Rule argument

Additional information

Used terms: a, hP, rs0, g, f(r; o)

Proof step 20 from proof mod-other-hp2 (previous step) (back to tree)

Lemma lift-aux

Proof step 21 from proof mod-other-hp2 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  rs0 ∩ rs2 = ∅, rs0 ∩ rs = ∅, rs ∩ rs2 = ∅, g(r) ≠ a0, g(r) ≠ a, hR[rs2, f, g], a0[hP0][rs, f, g], hP(mk(rs0, f)), ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs ∪ rs2, f(r; o), g], ∀ r. r ∈ rs0 → g(r) = a, ¬ hP(mk(rs0, f(r; o))) ∨ ¬(r0 ∈ rs0 → g(r0) = a) ⊦

Rule argument

Additional information

Proof step 22 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 23:

  rs ∩ rs2 = ∅, rs0 ∩ rs = ∅, rs0 ∩ rs2 = ∅, g(r) ≠ a0, g(r) ≠ a, a0[hP0][rs, f, g], hR[rs2, f, g], hP(mk(rs0, f)), ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs ∪ rs2, f(r; o), g], r ∈ rs0 → g(r) = a, ∀ r. r ∈ rs0 → g(r) = a, ¬ hP(mk(rs0, f(r; o))) ∨ ¬(r0 ∈ rs0 → g(r0) = a) ⊦

Rule argument

Additional information

Used terms: r

Proof step 23 from proof mod-other-hp2 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  rs0 ∩ rs2 = ∅, rs0 ∩ rs = ∅, rs ∩ rs2 = ∅, g(r) ≠ a0, g(r) ≠ a, hR[rs2, f, g], a0[hP0][rs, f, g], hP(mk(rs0, f)), ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs ∪ rs2, f(r; o), g], ¬ r ∈ rs0, ∀ r. r ∈ rs0 → g(r) = a, ¬ hP(mk(rs0, f(r; o))) ∨ ¬(r0 ∈ rs0 → g(r0) = a) ⊦

Rule argument

Additional information

Proof step 24 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 25:

  mk(rs0, f(r; o)) = mk(rs0, f), rs ∩ rs2 = ∅, rs0 ∩ rs = ∅, rs0 ∩ rs2 = ∅, g(r) ≠ a0, g(r) ≠ a, a0[hP0][rs, f, g], hR[rs2, f, g], hP(mk(rs0, f)), ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs ∪ rs2, f(r; o), g], ¬ r ∈ rs0, ∀ r. r ∈ rs0 → g(r) = a, ¬ hP(mk(rs0, f(r; o))) ∨ ¬(r0 ∈ rs0 → g(r0) = a) ⊦

• Proof step 27:

  rs ∩ rs2 = ∅, rs0 ∩ rs = ∅, rs0 ∩ rs2 = ∅, g(r) ≠ a0, g(r) ≠ a, mk(rs0, f(r; o)) ≠ mk(rs0, f), a0[hP0][rs, f, g], hR[rs2, f, g], hP(mk(rs0, f)), ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs ∪ rs2, f(r; o), g], ¬ r ∈ rs0, ∀ r. r ∈ rs0 → g(r) = a, ¬ hP(mk(rs0, f(r; o))) ∨ ¬(r0 ∈ rs0 → g(r0) = a) ⊦

Rule argument

Additional information

Used formulas: mk(rs0, f(r; o)) = mk(rs0, f)

Proof step 25 from proof mod-other-hp2 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  mk(rs0, f(r; o)) = mk(rs0, f), rs ∩ rs2 = ∅, rs0 ∩ rs = ∅, rs0 ∩ rs2 = ∅, g(r) ≠ a, g(r) ≠ a0, g(r0) ≠ a, a0[hP0][rs, f, g], hR[rs2, f, g], hP(mk(rs0, f)), r0 ∈ rs0, ¬ (a[hP] * a0[hP0] * true)[rs0 ∪ rs ∪ rs2, f(r; o), g], ¬ r ∈ rs0, ∀ r. r ∈ rs0 → g(r) = a ⊦

Rule argument

Additional information

Proof step 26 from proof mod-other-hp2 (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 27 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 28:

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

Rule argument

Additional information

Used simplifier rules: ⊦ r ∈ mk(rs, f) ↔ r ∈ rs; (s) ⊦ r ∈ rs → mk(rs, f)[r] = f(r); (s) Used terms: mk(rs0, f), mk(rs0, f(r; o))

Proof step 28 from proof mod-other-hp2 (previous step) (back to tree)

A spec-lemma

Proof step 29 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 30:

  ⊦ g(r) ≠ a ∧ (a[hP] * hR)[rs, f, g] → (a[hP] * true)[rs, f(r; o), g]

• Proof step 31:

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

Rule argument

Additional information

Used terms: true, r, a0, hP0, rs ∪ rs2, g, f, o

Proof step 30 from proof mod-other-hp2 (previous step) (back to tree)

Lemma mod-other-hp

Proof step 31 from proof mod-other-hp2 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

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

Rule argument

Additional information

Proof step 32 from proof mod-other-hp2 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 33:

  ⊦ 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: rs, true, g, a0[hP0], f, rs2

Proof step 33 from proof mod-other-hp2 (previous step) (back to tree)

A spec-lemma