Proof step 1 from proof in-uni (conclusion) (back to tree)

⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1)

System applied rule normalize

And obtained the following new goals

Proof step 2:
¬(r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1), h0 ⊥ h1 ⊦

Rule argument

Additional information

Proof step 2 from proof in-uni (previous step) (back to tree)

¬(r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1), h0 ⊥ h1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
¬(r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1), h0 ⊥ h1 ⊦

Rule argument

Additional information

Proof step 3 from proof in-uni (previous step) (back to tree)

¬(r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1), h0 ⊥ h1 ⊦

User applied rule induction

And obtained the following new goals

Proof step 4:
¬(r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1), h0 ⊥ h1, ∀ h, r, h0. # h < # h1 → ¬(r ∈ h0 ∪ h ↔ r ∈ h0 ∨ r ∈ h) ∧ h0 ⊥ h → false ⊦

Rule argument

indvar: # h1 wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: # h1

Proof step 4 from proof in-uni (previous step) (back to tree)

¬(r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1), h0 ⊥ h1, ∀ h, r, h0. # h < # h1 → ¬(r ∈ h0 ∪ h ↔ r ∈ h0 ∨ r ∈ h) ∧ h0 ⊥ h → false ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 5:
h1 ≠ ∅, ¬(r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1), h0 ⊥ h1, ∀ h, r, h0. # h < # h1 → ¬(r ∈ h0 ∪ h ↔ r ∈ h0 ∨ r ∈ h) ∧ h0 ⊥ h → false ⊦

Rule argument

fma: h1 = ∅

Additional information

Used simplifier rules: uni-emp: ⊦ h0 ∪ ∅ = h0; ⊦ ¬ a ∈ ∅; (s) Used formulas: h1 = ∅

Proof step 5 from proof in-uni (previous step) (back to tree)

h1 ≠ ∅, ¬(r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1), h0 ⊥ h1, ∀ h, r, h0. # h < # h1 → ¬(r ∈ h0 ∪ h ↔ r ∈ h0 ∨ r ∈ h) ∧ h0 ⊥ h → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ st = ∅ ↔ ¬(∃ a, d, st0. st = st0(a := d) ∧ ¬ a ∈ st0)
Proof step 7:
¬(r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1), h0 ⊥ h1, ∃ a, d, st0. h1 = st0(a := d) ∧ ¬ a ∈ st0, ∀ h, r, h0. # h < # h1 → ¬(r ∈ h0 ∪ h ↔ r ∈ h0 ∨ r ∈ h) ∧ h0 ⊥ h → false ⊦

Rule argument

spec: map, name:exnotempty, seq: ⊦ st = ∅ ↔ ¬(∃ a, d, st0. st = st0(a := d) ∧ ¬ a ∈ st0), subst:[st] → [h1]; [A, B] → [A, B], lhs:h1 = ∅, <all paths>

Additional information

Used terms: h1

Proof step 6 from proof in-uni (previous step) (back to tree)

⊦ st = ∅ ↔ ¬(∃ a, d, st0. st = st0(a := d) ∧ ¬ a ∈ st0)

A spec-lemma

Proof step 7 from proof in-uni (previous step) (back to tree)

¬(r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1), h0 ⊥ h1, ∃ a, d, st0. h1 = st0(a := d) ∧ ¬ a ∈ st0, ∀ h, r, h0. # h < # h1 → ¬(r ∈ h0 ∪ h ↔ r ∈ h0 ∨ r ∈ h) ∧ h0 ⊥ h → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
¬(r ∈ h0 ∪ h(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), h0 ⊥ h(r0 := d), ¬ r0 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ st(b := d) ↔ a = b ∨ a ∈ st; (s)

Proof step 8 from proof in-uni (previous step) (back to tree)

¬(r ∈ h0 ∪ h(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), h0 ⊥ h(r0 := d), ¬ r0 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 9:
h0 ⊥ h1 ⊦ h0 ∪ h1(r := o) = (h0 ∪ h1)(r := o)
Proof step 10:
¬(r ∈ h0 ∪ h(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), h0 ⊥ h(r0 := d), ¬ h0 ⊥ h, ¬ r0 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦
Proof step 17:
¬(r ∈ (h0 ∪ h)(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), h0 ⊥ h(r0 := d), h0 ⊥ h, ¬ r0 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

Rule argument

spec: <toplevel>, name:uni-put, seq:h0 ⊥ h1 ⊦ h0 ∪ h1(r := o) = (h0 ∪ h1)(r := o), subst:[h0, h1, r, o] → [h0, h, r0, d]; [A, B] → [A, B], lhs:h0 ∪ h(r0 := d), <all paths>

Additional information

Used terms: h0, h, r0, d

Proof step 9 from proof in-uni (previous step) (back to tree)

h0 ⊥ h1 ⊦ h0 ∪ h1(r := o) = (h0 ∪ h1)(r := o)

Lemma uni-put

Proof step 10 from proof in-uni (previous step) (back to tree)

¬(r ∈ h0 ∪ h(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), h0 ⊥ h(r0 := d), ¬ h0 ⊥ h, ¬ r0 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ h0 ⊥ h1 ↔ (∀ r. ¬ r ∈ h0 ∨ ¬ r ∈ h1)
Proof step 12:
¬(r ∈ h0 ∪ h(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), ¬ h0 ⊥ h, ¬ r0 ∈ h, ∀ r. ¬ r ∈ h0 ∨ ¬ r ∈ h(r0 := d), ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

Rule argument

spec: <toplevel>, name:disj, seq: ⊦ h0 ⊥ h1 ↔ (∀ r. ¬ r ∈ h0 ∨ ¬ r ∈ h1), subst:[h0, h1] → [h0, h(r0 := d)]; [A, B] → [A, B], lhs:h0 ⊥ h(r0 := d), <all paths>

Additional information

Used terms: h0, h(r0 := d)

Proof step 11 from proof in-uni (previous step) (back to tree)

⊦ h0 ⊥ h1 ↔ (∀ r. ¬ r ∈ h0 ∨ ¬ r ∈ h1)

Lemma disj

Proof step 12 from proof in-uni (previous step) (back to tree)

¬(r ∈ h0 ∪ h(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), ¬ h0 ⊥ h, ¬ r0 ∈ h, ∀ r. ¬ r ∈ h0 ∨ ¬ r ∈ h(r0 := d), ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
¬(r ∈ h0 ∪ h(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), ¬ h0 ⊥ h, ¬ r0 ∈ h, ∀ r1. ¬ r1 ∈ h0 ∨ r1 ≠ r0 ∧ ¬ r1 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ st(b := d) ↔ a = b ∨ a ∈ st; (s)

Proof step 13 from proof in-uni (previous step) (back to tree)

¬(r ∈ h0 ∪ h(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), ¬ h0 ⊥ h, ¬ r0 ∈ h, ∀ r1. ¬ r1 ∈ h0 ∨ r1 ≠ r0 ∧ ¬ r1 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ h0 ⊥ h1 ↔ (∀ r. ¬ r ∈ h0 ∨ ¬ r ∈ h1)
Proof step 15:
¬(r ∈ h0 ∪ h(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), ¬ r0 ∈ h, ∃ r. ¬(¬ r ∈ h0 ∨ ¬ r ∈ h), ∀ r1. ¬ r1 ∈ h0 ∨ r1 ≠ r0 ∧ ¬ r1 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

Rule argument

spec: <toplevel>, name:disj, seq: ⊦ h0 ⊥ h1 ↔ (∀ r. ¬ r ∈ h0 ∨ ¬ r ∈ h1), subst:[h0, h1] → [h0, h]; [A, B] → [A, B], lhs:h0 ⊥ h, <all paths>

Additional information

Used terms: h0, h

Proof step 14 from proof in-uni (previous step) (back to tree)

⊦ h0 ⊥ h1 ↔ (∀ r. ¬ r ∈ h0 ∨ ¬ r ∈ h1)

Lemma disj

Proof step 15 from proof in-uni (previous step) (back to tree)

¬(r ∈ h0 ∪ h(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), ¬ r0 ∈ h, ∃ r. ¬(¬ r ∈ h0 ∨ ¬ r ∈ h), ∀ r1. ¬ r1 ∈ h0 ∨ r1 ≠ r0 ∧ ¬ r1 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
¬(r ∈ h0 ∪ h(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), r1 ∈ h0, r1 ∈ h, ¬ r0 ∈ h, ∀ r1. ¬ r1 ∈ h0 ∨ r1 ≠ r0 ∧ ¬ r1 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

Rule argument

Additional information

Proof step 16 from proof in-uni (previous step) (back to tree)

¬(r ∈ h0 ∪ h(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), r1 ∈ h0, r1 ∈ h, ¬ r0 ∈ h, ∀ r1. ¬ r1 ∈ h0 ∨ r1 ≠ r0 ∧ ¬ r1 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: r1

Additional information

Used terms: r1

Proof step 17 from proof in-uni (previous step) (back to tree)

¬(r ∈ (h0 ∪ h)(r0 := d) ↔ r ∈ h0 ∨ r = r0 ∨ r ∈ h), h0 ⊥ h(r0 := d), h0 ⊥ h, ¬ r0 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
¬(r ∈ h0 ∪ h ↔ r ∈ h0 ∨ r ∈ h), r ≠ r0, h0 ⊥ h, h0 ⊥ h(r0 := d), ¬ r0 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ st(b := d) ↔ a = b ∨ a ∈ st; (s)

Proof step 18 from proof in-uni (previous step) (back to tree)

¬(r ∈ h0 ∪ h ↔ r ∈ h0 ∨ r ∈ h), r ≠ r0, h0 ⊥ h, h0 ⊥ h(r0 := d), ¬ r0 ∈ h, ∀ h0, r, h2. # h0 < # h(r0 := d) → ¬(r ∈ h2 ∪ h0 ↔ r ∈ h2 ∨ r ∈ h0) ∧ h2 ⊥ h0 → false ⊦

User applied rule apply induction

And closed the goal!

Rule argument

subst: [h0, r, h2] → [h, r, h0]

Additional information

Used simplifier rules: ⊦ ¬ a ∈ st → # st(a := d) = # st + 1; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ n < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: h, r, h0

Proof Tree for in-uni (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.