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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

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

Rule argument

Additional information

Proof step 3 from proof mod-other-both (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:

  g(r) ≠ a, ¬ (a[hP] * true)[rs, f(r; o), g(r; a0)], ∃ 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 mod-other-both (previous step) (back to tree)

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

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

• Proof step 8:

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

Rule argument

Additional information

Used terms: rs0, true, g(r; a0), a[hP], f(r; o), rs1

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

A spec-lemma

Proof step 8 from proof mod-other-both (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

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

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

  ⊦ hR[rs, f, g] ↔ hR(mk(rs, f, g))

• Proof step 11:

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

Rule argument

Additional information

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

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

A spec-lemma

Proof step 11 from proof mod-other-both (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 12:

  ⊦ hR[rs, f, g] ↔ hR(mk(rs, f, g))

• Proof step 13:

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

Rule argument

Additional information

Used terms: g(r; a0), f(r; o), rs0, a[hP]

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

A spec-lemma

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

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

• Proof step 15:

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

Rule argument

Additional information

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

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

Lemma lift-aux

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 17:

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

• Proof step 18:

  rs0 ∩ rs1 = ∅, g(r) ≠ a, hR[rs1, f, g], hP(mk(rs0, f)), ¬ (a[hP] * true)[rs0 ∪ rs1, f(r; o), g(r; a0)], ∀ r. r ∈ rs0 → g(r) = a, ¬ hP(mk(rs0, f(r; o))) ∨ (∃ r0. ¬(r0 ∈ rs0 → g(r; a0)(r0) = a)) ⊦

Rule argument

Additional information

Used terms: a, hP, rs0, g(r; a0), f(r; o)

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

Lemma lift-aux

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  rs0 ∩ rs1 = ∅, g(r) ≠ a, hR[rs1, f, g], hP(mk(rs0, f)), ¬ (a[hP] * true)[rs0 ∪ rs1, f(r; o), g(r; a0)], ∀ r. r ∈ rs0 → g(r) = a, ¬ hP(mk(rs0, f(r; o))) ∨ ¬(r0 ∈ rs0 → (r = r0 ⊃ a0;g(r0)) = a) ⊦

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 20:

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

• Proof step 27:

  rs0 ∩ rs1 = ∅, (r = r0 ⊃ a0;g(r0)) ≠ a, g(r) ≠ a, hR[rs1, f, g], hP(mk(rs0, f)), r0 ∈ rs0, ¬ (a[hP] * true)[rs0 ∪ rs1, f(r; o), g(r; a0)], ∀ r. r ∈ rs0 → g(r) = a ⊦

Rule argument

Additional information

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

User applied rule cut formula

And obtained the following new goals

• Proof step 21:

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

• Proof step 23:

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

Rule argument

Additional information

Used formulas: r ∈ rs0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

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

Rule argument

Additional information

Proof step 22 from proof mod-other-both (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 23 from proof mod-other-both (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

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

Rule argument

Additional information

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

User applied rule cut formula

And obtained the following new goals

• Proof step 25:

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

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  ⊦ 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 26 from proof mod-other-both (previous step) (back to tree)

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  rs0 ∩ rs1 = ∅, (r = r0 ⊃ a0;g(r0)) ≠ a, g(r) ≠ a, hR[rs1, f, g], hP(mk(rs0, f)), r0 ∈ rs0, ¬ (a[hP] * true)[rs0 ∪ rs1, f(r; o), g(r; a0)], ∀ r. r ∈ rs0 → g(r) = a ⊦

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 29:

  r = r0, rs0 ∩ rs1 = ∅, a0 ≠ a, g(r) ≠ a, hR[rs1, f, g], hP(mk(rs0, f)), r0 ∈ rs0, ¬ (a[hP] * true)[rs0 ∪ rs1, f(r; o), g(r; a0)], ∀ r. r ∈ rs0 → g(r) = a ⊦

• Proof step 31:

  rs0 ∩ rs1 = ∅, g(r0) ≠ a, g(r) ≠ a, r ≠ r0, hR[rs1, f, g], hP(mk(rs0, f)), r0 ∈ rs0, ¬ (a[hP] * true)[rs0 ∪ rs1, f(r; o), g(r; a0)], ∀ r. r ∈ rs0 → g(r) = a ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  rs0 ∩ rs1 = ∅, a0 ≠ a, g(r0) ≠ a, hR[rs1, f, g], hP(mk(rs0, f)), r0 ∈ rs0, ¬ (a[hP] * true)[rs0 ∪ rs1, f(r0; o), g(r0; a0)], ∀ r. r ∈ rs0 → g(r) = a ⊦

Rule argument

Additional information

Proof step 30 from proof mod-other-both (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 31 from proof mod-other-both (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

  rs0 ∩ rs1 = ∅, g(r0) ≠ a, g(r) ≠ a, r ≠ r0, hR[rs1, f, g], hP(mk(rs0, f)), r0 ∈ rs0, ¬ (a[hP] * true)[rs0 ∪ rs1, f(r; o), g(r; a0)], ∀ r. r ∈ rs0 → g(r) = a ⊦

Rule argument

Additional information

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: r0