Proof step 1 from proof rm-allsize (conclusion) (back to tree)

∀ a. a ∈ x ↔ a ∈ y ⊦ # x.distinct = # y.distinct

System applied rule normalize

And obtained the following new goals

Proof step 2:
# x.distinct ≠ # y.distinct, ∀ a. a ∈ x ↔ a ∈ y ⊦

Rule argument

Additional information

Proof step 2 from proof rm-allsize (previous step) (back to tree)

# x.distinct ≠ # y.distinct, ∀ a. a ∈ x ↔ a ∈ y ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 3:
# [].distinct ≠ # y.distinct, ∀ a. a ∈ [] ↔ a ∈ y ⊦
Proof step 5:
# (a + x0).distinct ≠ # y.distinct, ∀ a0. a0 ∈ a + x0 ↔ a0 ∈ y, ∀ y. ¬(# x0.distinct ≠ # y.distinct ∧ (∀ a. a ∈ x0 ↔ a ∈ y)) ⊦

Rule argument

indvar: x

Additional information

Used simplifier rules: List[A] freely generated by [], +;

Proof step 3 from proof rm-allsize (previous step) (back to tree)

# [].distinct ≠ # y.distinct, ∀ a. a ∈ [] ↔ a ∈ y ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
y ≠ [], ∀ a. ¬ a ∈ y ⊦

Rule argument

Additional information

Used simplifier rules: rmdup-e: ⊦ [].distinct = []; rm-empty: ⊦ x.distinct = [] ↔ x = []; ⊦ ¬ a ∈ []; (s) ⊦ # x = 0 ↔ x = []; (s) ⊦ # [] = 0; (s)

Proof step 4 from proof rm-allsize (previous step) (back to tree)

y ≠ [], ∀ a. ¬ a ∈ y ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: y.head abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ x.head ∈ x ↔ x ≠ []; (s) Used terms: y.head

Proof step 5 from proof rm-allsize (previous step) (back to tree)

# (a + x0).distinct ≠ # y.distinct, ∀ a0. a0 ∈ a + x0 ↔ a0 ∈ y, ∀ y. ¬(# x0.distinct ≠ # y.distinct ∧ (∀ a. a ∈ x0 ↔ a ∈ y)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# (a ' + x0).distinct ≠ # y.distinct, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)

Proof step 6 from proof rm-allsize (previous step) (back to tree)

# (a ' + x0).distinct ≠ # y.distinct, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 7:
# (a ' + x0).distinct ≠ # y.distinct, a ∈ x0, ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦
Proof step 11:
# (a ' + x0).distinct ≠ # y.distinct, ¬ a ∈ x0, ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Rule argument

fma: a ∈ x0

Additional information

Used formulas: a ∈ x0

Proof step 7 from proof rm-allsize (previous step) (back to tree)

# (a ' + x0).distinct ≠ # y.distinct, a ∈ x0, ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# x0.distinct ≠ # y.distinct, a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y ⊦

Rule argument

Additional information

Used simplifier rules: rmdup-y: ⊦ a ∈ x → (a ' + x).distinct = x.distinct; a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 8 from proof rm-allsize (previous step) (back to tree)

# x0.distinct ≠ # y.distinct, a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y ⊦

User applied rule all left

And obtained the following new goals

Proof step 9:
# x0.distinct ≠ # y.distinct, a ∈ x0, ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), # x0.distinct = # y.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y) ⊦

Rule argument

pos: left 3 terms: y abort: F discard: F computed: T

Additional information

Used terms: y

Proof step 9 from proof rm-allsize (previous step) (back to tree)

# x0.distinct ≠ # y.distinct, a ∈ x0, ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), # x0.distinct = # y.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# x0.distinct ≠ # y.distinct, ¬(a0 ∈ x0 ↔ a0 ∈ y), a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a1. a1 ∈ x0 ↔ a1 ∈ y ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ a0 ' ↔ a = a0; (s) ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)

Proof step 10 from proof rm-allsize (previous step) (back to tree)

# x0.distinct ≠ # y.distinct, ¬(a0 ∈ x0 ↔ a0 ∈ y), a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a1. a1 ∈ x0 ↔ a1 ∈ y ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: a0 abort: F discard: F computed: T

Additional information

Used terms: a0

Proof step 11 from proof rm-allsize (previous step) (back to tree)

# (a ' + x0).distinct ≠ # y.distinct, ¬ a ∈ x0, ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
1 + # x0.distinct ≠ # y.distinct, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y ⊦

Rule argument

Additional information

Used simplifier rules: rmdup-n: ⊦ ¬ a ∈ x → (a ' + x).distinct = a + x.distinct; a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ # a ' = 0.+1; (s) ⊦ #(x + y) = # x + # y; (s) ⊦ a + x = a ' + x; (s)

Proof step 12 from proof rm-allsize (previous step) (back to tree)

1 + # x0.distinct ≠ # y.distinct, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y ⊦

User applied rule all left

And obtained the following new goals

Proof step 13:
a ∈ a ' + x0 ↔ a ∈ y, 1 + # x0.distinct ≠ # y.distinct, ¬ a ∈ x0, ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Rule argument

pos: left 4 terms: a abort: F discard: F computed: T

Additional information

Used terms: a

Proof step 13 from proof rm-allsize (previous step) (back to tree)

a ∈ a ' + x0 ↔ a ∈ y, 1 + # x0.distinct ≠ # y.distinct, ¬ a ∈ x0, ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
a ∈ a ' + x0 ↔ a ∈ y, 1 + # x0.distinct ≠ # y.distinct, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y ⊦

Rule argument

Additional information

Proof step 14 from proof rm-allsize (previous step) (back to tree)

a ∈ a ' + x0 ↔ a ∈ y, 1 + # x0.distinct ≠ # y.distinct, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 15:
⊦ a ∈ x → (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y)
Proof step 16:
a ∈ a ' + x0 ↔ a ∈ y, 1 + # x0.distinct ≠ # y.distinct, ¬ a ∈ x0, a ∈ y → (∃ y0, z. y = y0 + a ' + z ∧ ¬ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Rule argument

spec: list, name:firstoc, seq: ⊦ a ∈ x → (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y), subst:[a, x] → [a, y]

Additional information

Used terms: a, y

Proof step 15 from proof rm-allsize (previous step) (back to tree)

⊦ a ∈ x → (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y)

A spec-lemma

Proof step 16 from proof rm-allsize (previous step) (back to tree)

a ∈ a ' + x0 ↔ a ∈ y, 1 + # x0.distinct ≠ # y.distinct, ¬ a ∈ x0, a ∈ y → (∃ y0, z. y = y0 + a ' + z ∧ ¬ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
1 + # x0.distinct ≠ # (y0 + a ' + z).distinct, ¬ a ∈ y0, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ a0 ' ↔ a = a0; (s) ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s) in-13: ⊦ x = y + a ' + z → a ∈ x; (forward) from specification list ⊦ x = y + a ' + z → a ∈ x; (s)

Proof step 17 from proof rm-allsize (previous step) (back to tree)

1 + # x0.distinct ≠ # (y0 + a ' + z).distinct, ¬ a ∈ y0, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 18:
⊦ ¬ a ∈ x ∧ ¬ a ∈ y → # (x + a ' + y).distinct = # (x + y).distinct + 1
Proof step 19:
1 + # x0.distinct ≠ # (y0 + a ' + z).distinct, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦ ¬ a ∈ y0 ∧ ¬ a ∈ z
Proof step 42:
1 + # x0.distinct ≠ # (y0 + z).distinct + 1, ¬ a ∈ z, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Rule argument

spec: <toplevel>, name: rm-size, seq: ⊦ ¬ a ∈ x ∧ ¬ a ∈ y → # (x + a ' + y).distinct = # (x + y).distinct + 1, subst:[y, a, x] → [z, a, y0], <all paths>

Additional information

Used terms: z, a, y0

Proof step 18 from proof rm-allsize (previous step) (back to tree)

⊦ ¬ a ∈ x ∧ ¬ a ∈ y → # (x + a ' + y).distinct = # (x + y).distinct + 1

Lemma rm-size

Proof step 19 from proof rm-allsize (previous step) (back to tree)

1 + # x0.distinct ≠ # (y0 + a ' + z).distinct, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦ ¬ a ∈ y0 ∧ ¬ a ∈ z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
1 + # x0.distinct ≠ # (y0 + z).distinct, a ∈ z, ¬ a ∈ y0, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

Rule argument

Additional information

Used simplifier rules: rm-size-02: ⊦ a ∈ y → # (x + a ' + y).distinct = # (x + y).distinct; a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 20 from proof rm-allsize (previous step) (back to tree)

1 + # x0.distinct ≠ # (y0 + z).distinct, a ∈ z, ¬ a ∈ y0, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 21:
⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z))
Proof step 22:
1 + # x0.distinct ≠ # (y0 + z).distinct, a ∈ z, ¬ a ∈ y0, ¬ a ∈ x0, a ∈ z.distinct ↔ (∃ y, z0. z.distinct = y + (a + z0)), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Rule argument

spec: list, name:in-def, seq: ⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z)), subst:[a, x] → [a, z.distinct]

Additional information

Used terms: a, z.distinct

Proof step 21 from proof rm-allsize (previous step) (back to tree)

⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z))

A spec-lemma

Proof step 22 from proof rm-allsize (previous step) (back to tree)

1 + # x0.distinct ≠ # (y0 + z).distinct, a ∈ z, ¬ a ∈ y0, ¬ a ∈ x0, a ∈ z.distinct ↔ (∃ y, z0. z.distinct = y + (a + z0)), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
z.distinct = y + a ' + z0, 1 + # x0.distinct ≠ # (y0 + z).distinct, a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

Rule argument

Additional information

Used simplifier rules: rm-in-03: ⊦ ¬ a ∈ x.distinct → ¬ a ∈ x; rm-notin-01: ⊦ x.distinct = y + a ' + z → a ∈ x ∧ ¬ a ∈ y ∧ ¬ a ∈ z; ⊦ x = y + a ' + z → a ∈ x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list in-13: ⊦ x = y + a ' + z → a ∈ x; (forward) from specification list ⊦ ¬ a ∈ x → x ≠ y + a ' + z; (s) ⊦ a + x = a ' + x; (s)

Proof step 23 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, 1 + # x0.distinct ≠ # (y0 + z).distinct, a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 24:
⊦ (x + y.distinct).distinct = (x + y).distinct
Proof step 25:
z.distinct = y + a ' + z0, 1 + # x0.distinct ≠ # (y0 + z.distinct).distinct, a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Rule argument

spec: <toplevel>, name: rm-rm-02, seq: ⊦ (x + y.distinct).distinct = (x + y).distinct, subst:[y, x] → [z, y0], <rotated> <all paths>

Additional information

Used terms: z, y0

Proof step 24 from proof rm-allsize (previous step) (back to tree)

⊦ (x + y.distinct).distinct = (x + y).distinct

Lemma rm-rm-02

Proof step 25 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, 1 + # x0.distinct ≠ # (y0 + z.distinct).distinct, a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

User applied rule subst equation

And obtained the following new goals

Proof step 26:
z.distinct = y + a ' + z0, 1 + # x0.distinct ≠ # (y0 + y + a ' + z0).distinct, a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

Rule argument

pos: left 1, flags:

Additional information

Proof step 26 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, 1 + # x0.distinct ≠ # (y0 + y + a ' + z0).distinct, a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ ¬ a ∈ x ∧ ¬ a ∈ y → # (x + a ' + y).distinct = # (x + y).distinct + 1
Proof step 28:
z.distinct = y + a ' + z0, 1 + # x0.distinct ≠ # ((y0 + y) + z0).distinct + 1, a ∈ z, ¬ a ∈ y0 + y, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Rule argument

spec: <toplevel>, name: rm-size, seq: ⊦ ¬ a ∈ x ∧ ¬ a ∈ y → # (x + a ' + y).distinct = # (x + y).distinct + 1, subst:[y, a, x] → [z0, a, y0 + y], <all paths>

Additional information

Used simplifier rules: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: z0, a, y0 + y

Proof step 27 from proof rm-allsize (previous step) (back to tree)

⊦ ¬ a ∈ x ∧ ¬ a ∈ y → # (x + a ' + y).distinct = # (x + y).distinct + 1

Lemma rm-size

Proof step 28 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, 1 + # x0.distinct ≠ # ((y0 + y) + z0).distinct + 1, a ∈ z, ¬ a ∈ y0 + y, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 29 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

User applied rule all left

And obtained the following new goals

Proof step 30:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), # x0.distinct = # ((y0 + y) + z0).distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ (y0 + y) + z0) ⊦

Rule argument

pos: left 8 terms: (y0 + y) + z0 abort: F discard: T computed: T

Additional information

Used terms: (y0 + y) + z0

Proof step 30 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), # x0.distinct = # ((y0 + y) + z0).distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ (y0 + y) + z0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, ¬(a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ y ∨ a0 ∈ z0), a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)

Proof step 31 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, ¬(a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ y ∨ a0 ∈ z0), a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

User applied rule all left

And obtained the following new goals

Proof step 32:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, ¬(a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ y ∨ a0 ∈ z0), a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

Rule argument

pos: left 9 terms: a0 abort: F discard: T computed: T

Additional information

Used terms: a0

Proof step 32 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, ¬(a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ y ∨ a0 ∈ z0), a ∈ z, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 33:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, a0 ∈ x0, ¬ a0 ∈ y0, ¬ a0 ∈ y, ¬ a0 ∈ z0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦
Proof step 37:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a0 ∈ x0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), a0 ∈ y0 ∨ a0 ∈ y ∨ a0 ∈ z0 ⊦

Rule argument

pos: left 3

Additional information

Proof step 33 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, a0 ∈ x0, ¬ a0 ∈ y0, ¬ a0 ∈ y, ¬ a0 ∈ z0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a0 ∈ x0, a ∈ z, a0 ∈ z, ¬ a0 ∈ y0, ¬ a0 ∈ y, ¬ a0 ∈ z0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0 ⊦

Rule argument

Additional information

Proof step 34 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a0 ∈ x0, a ∈ z, a0 ∈ z, ¬ a0 ∈ y0, ¬ a0 ∈ y, ¬ a0 ∈ z0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0 ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 35:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a0 ∈ z, a ∈ z, a0 ∈ x0, a0 ∈ z.distinct, ¬ a0 ∈ y0, ¬ a0 ∈ y, ¬ a0 ∈ z0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0 ⊦

Rule argument

fma: a0 ∈ z.distinct

Additional information

Used simplifier rules: rm-in-03: ⊦ ¬ a ∈ x.distinct → ¬ a ∈ x; Used formulas: a0 ∈ z.distinct

Proof step 35 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a0 ∈ z, a ∈ z, a0 ∈ x0, a0 ∈ z.distinct, ¬ a0 ∈ y0, ¬ a0 ∈ y, ¬ a0 ∈ z0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0 ⊦

User applied rule subst equation

And obtained the following new goals

Proof step 36:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a0 ∈ y + a ' + z0, a0 ∈ x0, a ∈ z, a0 ∈ z, ¬ a0 ∈ y0, ¬ a0 ∈ y, ¬ a0 ∈ z0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0 ⊦

Rule argument

pos: left 1, flags:

Additional information

Proof step 36 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a0 ∈ y + a ' + z0, a0 ∈ x0, a ∈ z, a0 ∈ z, ¬ a0 ∈ y0, ¬ a0 ∈ y, ¬ a0 ∈ z0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0 ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ ¬ a ∈ x → (a ∈ a0 ' + x ↔ a = a0); (s) ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)

Proof step 37 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a0 ∈ x0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), a0 ∈ y0 ∨ a0 ∈ y ∨ a0 ∈ z0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a0 ∈ x0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ≠ a → ¬ a0 ∈ y0 ∧ ¬ a0 ∈ z, a0 ∈ y0 ∨ a0 ∈ y ∨ a0 ∈ z0 ⊦

Rule argument

Additional information

Proof step 38 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a0 ∈ x0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ≠ a → ¬ a0 ∈ y0 ∧ ¬ a0 ∈ z, a0 ∈ y0 ∨ a0 ∈ y ∨ a0 ∈ z0 ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 39:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a0 ∈ z, ¬ a0 ∈ y0, ¬ a0 ∈ x0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ∈ y0 ∨ a0 ∈ y ∨ a0 ∈ z0 ⊦

Rule argument

pos: left 9

Additional information

Proof step 39 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a0 ∈ z, ¬ a0 ∈ y0, ¬ a0 ∈ x0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ∈ y0 ∨ a0 ∈ y ∨ a0 ∈ z0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a0 ∈ z, ¬ a0 ∈ y0, ¬ a0 ∈ x0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ∈ y ∨ a0 ∈ z0 ⊦

Rule argument

Additional information

Proof step 40 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a0 ∈ z, ¬ a0 ∈ y0, ¬ a0 ∈ x0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ∈ y ∨ a0 ∈ z0 ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 41:
z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a0 ∈ z.distinct, ¬ a0 ∈ z, ¬ a0 ∈ y0, ¬ a0 ∈ x0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ∈ y ∨ a0 ∈ z0 ⊦

Rule argument

fma: a0 ∈ z.distinct

Additional information

Used simplifier rules: rm-in-01: ⊦ a ∈ x.distinct → a ∈ x; Used formulas: a0 ∈ z.distinct

Proof step 41 from proof rm-allsize (previous step) (back to tree)

z.distinct = y + a ' + z0, # x0.distinct ≠ # ((y0 + y) + z0).distinct, a ∈ z, ¬ a0 ∈ z.distinct, ¬ a0 ∈ z, ¬ a0 ∈ y0, ¬ a0 ∈ x0, ¬ a ∈ z0, ¬ a ∈ y, ¬ a ∈ y0, ¬ a ∈ x0, a0 ∈ y ∨ a0 ∈ z0 ⊦

User applied rule subst equation

And closed the goal!

Rule argument

pos: left 1, flags: discard

Additional information

Used simplifier rules: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s) a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 42 from proof rm-allsize (previous step) (back to tree)

1 + # x0.distinct ≠ # (y0 + z).distinct + 1, ¬ a ∈ z, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
# x0.distinct ≠ # (y0 + z).distinct, ¬ a ∈ z, ¬ a ∈ y0, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 43 from proof rm-allsize (previous step) (back to tree)

# x0.distinct ≠ # (y0 + z).distinct, ¬ a ∈ z, ¬ a ∈ y0, ¬ a ∈ x0, ∀ y0. # x0.distinct = # y0.distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

User applied rule all left

And obtained the following new goals

Proof step 44:
# x0.distinct ≠ # (y0 + z).distinct, ¬ a ∈ z, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), # x0.distinct = # (y0 + z).distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0 + z) ⊦

Rule argument

pos: left 5 terms: y0 + z abort: F discard: T computed: T

Additional information

Used terms: y0 + z

Proof step 44 from proof rm-allsize (previous step) (back to tree)

# x0.distinct ≠ # (y0 + z).distinct, ¬ a ∈ z, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), # x0.distinct = # (y0 + z).distinct ∨ ¬(∀ a. a ∈ x0 ↔ a ∈ y0 + z) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 45:
# x0.distinct ≠ # (y0 + z).distinct, ¬(a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), ¬ a ∈ z, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)

Proof step 45 from proof rm-allsize (previous step) (back to tree)

# x0.distinct ≠ # (y0 + z).distinct, ¬(a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z), ¬ a ∈ z, ¬ a ∈ y0, ¬ a ∈ x0, ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: a0 abort: F discard: T computed: T

Additional information

Used terms: a0

Proof Tree for rm-allsize (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.