Proof step 1 from proof rm-size-01 (conclusion) (back to tree)

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

System applied rule normalize

And obtained the following new goals

Proof step 2:
# (x + a ' + y).distinct ≠ # (x + y).distinct, a ∈ x ⊦

Rule argument

Additional information

Proof step 2 from proof rm-size-01 (previous step) (back to tree)

# (x + a ' + y).distinct ≠ # (x + y).distinct, a ∈ x ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# (x + a ' + y).distinct ≠ # (x + y).distinct, a ∈ x ⊦

Rule argument

Additional information

Proof step 3 from proof rm-size-01 (previous step) (back to tree)

# (x + a ' + y).distinct ≠ # (x + y).distinct, a ∈ x ⊦

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
# ((a0 + x0) + a ' + y).distinct ≠ # ((a0 + x0) + y).distinct, a ∈ a0 + x0, ∀ a, y. ¬(# (x0 + a ' + y).distinct ≠ # (x0 + y).distinct ∧ a ∈ x0) ⊦

Rule argument

term: x

Additional information

Used simplifier rules: List('a) freely generated by [], +;; ⊦ [] + x = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ ¬ a ∈ []; (s)

Proof step 4 from proof rm-size-01 (previous step) (back to tree)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# ((a0 ' + x0) + a ' + y).distinct ≠ # ((a0 ' + x0) + y).distinct, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0, a = a0 ∨ a ∈ x0 ⊦

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) ⊦ a + x = a ' + x; (s)

Proof step 5 from proof rm-size-01 (previous step) (back to tree)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
# ((a0 ' + x0) + a ' + y).distinct ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0, a = a0 ∨ a ∈ x0 ⊦
Proof step 22:
# ((a0 ' + x0) + a ' + y).distinct ≠ # ((a0 ' + x0) + y).distinct, ¬ a ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0, a = a0 ∨ a ∈ x0 ⊦

Rule argument

fma: a ∈ x0

Additional information

Used formulas: a ∈ x0

Proof step 6 from proof rm-size-01 (previous step) (back to tree)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# ((a0 ' + x0) + a ' + y).distinct ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Rule argument

Additional information

Proof step 7 from proof rm-size-01 (previous step) (back to tree)

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

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ (a ' + x).distinct = (a ∈ x ⊃ x.distinct;a ' + x.distinct)
Proof step 9:
# (a0 ∈ x0 + a ' + y ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Rule argument

spec: <toplevel>, name: rm-ifrec, seq: ⊦ (a ' + x).distinct = (a ∈ x ⊃ x.distinct;a ' + x.distinct), subst:[x, a] → [x0 + a ' + y, a0], <all paths>

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0 + a ' + y, a0

Proof step 8 from proof rm-size-01 (previous step) (back to tree)

⊦ (a ' + x).distinct = (a ∈ x ⊃ x.distinct;a ' + x.distinct)

Lemma rm-ifrec

Proof step 9 from proof rm-size-01 (previous step) (back to tree)

# (a0 ∈ x0 + a ' + y ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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-size-01 (previous step) (back to tree)

# (a0 ∈ x0 ∨ a0 ∈ y ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 11:
# (a0 ∈ x0 ∨ a0 ∈ y ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦
Proof step 13:
# (a0 ∈ x0 ∨ a0 ∈ y ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Rule argument

fma: a0 ∈ y

Additional information

Used formulas: a0 ∈ y

Proof step 11 from proof rm-size-01 (previous step) (back to tree)

# (a0 ∈ x0 ∨ a0 ∈ y ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# (x0 + a ' + y).distinct ≠ # (x0 + y).distinct, a0 ∈ y, a ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Rule argument

Additional information

Used simplifier rules: rm-in-04: ⊦ a ∈ y → (a ' + x + y).distinct = (x + y).distinct; a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 12 from proof rm-size-01 (previous step) (back to tree)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a, y

Additional information

Used terms: a, y

Proof step 13 from proof rm-size-01 (previous step) (back to tree)

# (a0 ∈ x0 ∨ a0 ∈ y ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# (a0 ∈ x0 ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Rule argument

Additional information

Proof step 14 from proof rm-size-01 (previous step) (back to tree)

# (a0 ∈ x0 ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 15:
# (a0 ∈ x0 ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, a0 ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦
Proof step 17:
# (a0 ∈ x0 ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, ¬ a0 ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Rule argument

fma: a0 ∈ x0

Additional information

Used formulas: a0 ∈ x0

Proof step 15 from proof rm-size-01 (previous step) (back to tree)

# (a0 ∈ x0 ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, a0 ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# (x0 + a ' + y).distinct ≠ # (x0 + y).distinct, a0 ∈ x0, a ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Rule argument

Additional information

Used simplifier rules: rm-in-05: ⊦ a ∈ x → (a ' + x + y).distinct = (x + y).distinct; a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 16 from proof rm-size-01 (previous step) (back to tree)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: a, y

Additional information

Used terms: a, y

Proof step 17 from proof rm-size-01 (previous step) (back to tree)

# (a0 ∈ x0 ⊃ (x0 + a ' + y).distinct;a0 ' + (x0 + a ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, ¬ a0 ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
1 + # (x0 + a ' + y).distinct ≠ # ((a0 ' + x0) + y).distinct, a ∈ x0, ¬ a0 ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ # a ' = 0 +1; (s) ⊦ #(x + y) = # x + # y; (s)

Proof step 18 from proof rm-size-01 (previous step) (back to tree)

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

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ (a ' + x).distinct = (a ∈ x ⊃ x.distinct;a ' + x.distinct)
Proof step 20:
1 + # (x0 + a ' + y).distinct ≠ # (a0 ∈ x0 + y ⊃ (x0 + y).distinct;a0 ' + (x0 + y).distinct), a ∈ x0, ¬ a0 ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Rule argument

spec: <toplevel>, name: rm-ifrec, seq: ⊦ (a ' + x).distinct = (a ∈ x ⊃ x.distinct;a ' + x.distinct), subst:[x, a] → [x0 + y, a0], <all paths>

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0 + y, a0

Proof step 19 from proof rm-size-01 (previous step) (back to tree)

⊦ (a ' + x).distinct = (a ∈ x ⊃ x.distinct;a ' + x.distinct)

Lemma rm-ifrec

Proof step 20 from proof rm-size-01 (previous step) (back to tree)

1 + # (x0 + a ' + y).distinct ≠ # (a0 ∈ x0 + y ⊃ (x0 + y).distinct;a0 ' + (x0 + y).distinct), a ∈ x0, ¬ a0 ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
# (x0 + a ' + y).distinct ≠ # (x0 + y).distinct, a ∈ x0, ¬ a0 ∈ x0, ¬ a0 ∈ y, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

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) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ # a ' = 0 +1; (s) ⊦ #(x + y) = # x + # y; (s) ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)

Proof step 21 from proof rm-size-01 (previous step) (back to tree)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: a, y

Additional information

Used terms: a, y

Proof step 22 from proof rm-size-01 (previous step) (back to tree)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
# ((a0 ' + x0) + a0 ' + y).distinct ≠ # ((a0 ' + x0) + y).distinct, ¬ a0 ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Rule argument

Additional information

Proof step 23 from proof rm-size-01 (previous step) (back to tree)

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

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 24:
⊦ (a ' + x).distinct = (a ∈ x ⊃ x.distinct;a ' + x.distinct)
Proof step 25:
# (a0 ∈ x0 + a0 ' + y ⊃ (x0 + a0 ' + y).distinct;a0 ' + (x0 + a0 ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, ¬ a0 ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Rule argument

spec: <toplevel>, name: rm-ifrec, seq: ⊦ (a ' + x).distinct = (a ∈ x ⊃ x.distinct;a ' + x.distinct), subst:[x, a] → [x0 + a0 ' + y, a0], <all paths>

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0 + a0 ' + y, a0

Proof step 24 from proof rm-size-01 (previous step) (back to tree)

⊦ (a ' + x).distinct = (a ∈ x ⊃ x.distinct;a ' + x.distinct)

Lemma rm-ifrec

Proof step 25 from proof rm-size-01 (previous step) (back to tree)

# (a0 ∈ x0 + a0 ' + y ⊃ (x0 + a0 ' + y).distinct;a0 ' + (x0 + a0 ' + y).distinct) ≠ # ((a0 ' + x0) + y).distinct, ¬ a0 ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
# (x0 + a0 ' + y).distinct ≠ # ((a0 ' + x0) + y).distinct, ¬ a0 ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

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 26 from proof rm-size-01 (previous step) (back to tree)

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

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ (a ' + x).distinct = (a ∈ x ⊃ x.distinct;a ' + x.distinct)
Proof step 28:
# (x0 + a0 ' + y).distinct ≠ # (a0 ∈ x0 + y ⊃ (x0 + y).distinct;a0 ' + (x0 + y).distinct), ¬ a0 ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Rule argument

spec: <toplevel>, name: rm-ifrec, seq: ⊦ (a ' + x).distinct = (a ∈ x ⊃ x.distinct;a ' + x.distinct), subst:[x, a] → [x0 + y, a0], <all paths>

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0 + y, a0

Proof step 27 from proof rm-size-01 (previous step) (back to tree)

⊦ (a ' + x).distinct = (a ∈ x ⊃ x.distinct;a ' + x.distinct)

Lemma rm-ifrec

Proof step 28 from proof rm-size-01 (previous step) (back to tree)

# (x0 + a0 ' + y).distinct ≠ # (a0 ∈ x0 + y ⊃ (x0 + y).distinct;a0 ' + (x0 + y).distinct), ¬ a0 ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
# (x0 + a0 ' + y).distinct ≠ # (a0 ∈ y ⊃ (x0 + y).distinct;a0 ' + (x0 + y).distinct), ¬ a0 ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

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 29 from proof rm-size-01 (previous step) (back to tree)

# (x0 + a0 ' + y).distinct ≠ # (a0 ∈ y ⊃ (x0 + y).distinct;a0 ' + (x0 + y).distinct), ¬ a0 ∈ x0, ∀ a1, y0. # (x0 + a1 ' + y0).distinct = # (x0 + y0).distinct ∨ ¬ a1 ∈ x0 ⊦

Heuristic cut applied rule cut formula

And closed the goal!

Rule argument

fma: a0 ∈ y

Additional information

Used simplifier rules: rm-size-02: ⊦ a ∈ y → # (x + a ' + y).distinct = # (x + y).distinct; rm-size: ⊦ ¬ a ∈ x ∧ ¬ a ∈ y → # (x + a ' + y).distinct = # (x + y).distinct + 1; ⊦ #(x + y) = # x + # y; (s) ⊦ # a ' = 0 +1; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m + n = m + n0 ↔ n = n0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used formulas: a0 ∈ y

Proof Tree for rm-size-01 (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.