Proof step 1 from proof delapp (conclusion) (back to tree)

⊦ perm(x + y, y -1l a) ↔ x = [] ∧ ¬ a ∈ y

System applied rule normalize

And obtained the following new goals

Proof step 2:
¬(perm(x + y, y -1l a) ↔ x = [] ∧ ¬ a ∈ y) ⊦

Rule argument

Additional information

Proof step 2 from proof delapp (previous step) (back to tree)

¬(perm(x + y, y -1l a) ↔ x = [] ∧ ¬ a ∈ y) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ (∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a)) ∧ (∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y)) ↔ x = [] ∧ ¬ a ∈ y

Rule argument

Additional information

Used simplifier rules: perm: ⊦ perm(x, y) ↔ x ⊆m y ∧ y ⊆m x; msubset: ⊦ x ⊆m y ↔ (∀ a. #oc(a, x) ≤ #oc(a, y)); ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y); a: ⊦ (x + y) + z = x + y + z; from specification polylist

Proof step 3 from proof delapp (previous step) (back to tree)

⊦ (∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a)) ∧ (∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y)) ↔ x = [] ∧ ¬ a ∈ y

User applied rule cut formula

And obtained the following new goals

Proof step 4:
x = [] ⊦ (∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a)) ∧ (∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y)) ↔ x = [] ∧ ¬ a ∈ y
Proof step 8:
x ≠ [] ⊦ (∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a)) ∧ (∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y)) ↔ x = [] ∧ ¬ a ∈ y

Rule argument

fma: x = []

Additional information

Used formulas: x = []

Proof step 4 from proof delapp (previous step) (back to tree)

x = [] ⊦ (∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a)) ∧ (∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y)) ↔ x = [] ∧ ¬ a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ (∀ a0. #oc(a0, y) ≤ #oc(a0, y -1l a)) ∧ (∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, y)) ↔ ¬ a ∈ y

Rule argument

Additional information

Used simplifier rules: oc-e: ⊦ #oc(a, []) = 0; c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ 0 + m = m; (s)

Proof step 5 from proof delapp (previous step) (back to tree)

⊦ (∀ a0. #oc(a0, y) ≤ #oc(a0, y -1l a)) ∧ (∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, y)) ↔ ¬ a ∈ y

User applied rule case distinction

And obtained the following new goals

Proof step 6:
a ∈ y, ∀ a0. #oc(a0, y) ≤ #oc(a0, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, y) ⊦

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ ¬ a ∈ x → x -1l a = x; (s) ⊦ m ≤ m; (s)

Proof step 6 from proof delapp (previous step) (back to tree)

a ∈ y, ∀ a0. #oc(a0, y) ≤ #oc(a0, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, y) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a ∈ y, ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, y), ∀ a0. #oc(a0, y) ≤ #oc(a0, y -1l a) ⊦

Rule argument

Additional information

Proof step 7 from proof delapp (previous step) (back to tree)

a ∈ y, ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, y), ∀ a0. #oc(a0, y) ≤ #oc(a0, y -1l a) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a

Additional information

Used simplifier rules: del1: ⊦ a ∈ x → #oc(a, x -1l a) = #oc(a, x) -1; oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x; ⊦ n -1 = n - 1; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < 1 ↔ m = 0; (s) n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s) ⊦ m + m0 ≤ m ↔ m0 = 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: a

Proof step 8 from proof delapp (previous step) (back to tree)

x ≠ [] ⊦ (∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a)) ∧ (∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y)) ↔ x = [] ∧ ¬ a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
x ≠ [], ∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y) ⊦

Rule argument

Additional information

Proof step 9 from proof delapp (previous step) (back to tree)

x ≠ [], ∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y) ⊦

User applied rule all left

And obtained the following new goals

Proof step 10:
x ≠ [], #oc(x.head, x) + #oc(x.head, y) ≤ #oc(x.head, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y), ∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a) ⊦

Rule argument

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

Additional information

Used terms: x.head

Proof step 10 from proof delapp (previous step) (back to tree)

x ≠ [], #oc(x.head, x) + #oc(x.head, y) ≤ #oc(x.head, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y), ∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a) ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Proof step 11:
x = a1 + y0, x ≠ [], #oc(a1, x) + #oc(a1, y) ≤ #oc(a1, y -1l a), ∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y) ⊦

Rule argument

spec: polylist-data, name:elim-list-c, seq:x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y, subst:[x, a, y] → [x, a1, y0]

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x, a1, y0

Proof step 11 from proof delapp (previous step) (back to tree)

x = a1 + y0, x ≠ [], #oc(a1, x) + #oc(a1, y) ≤ #oc(a1, y -1l a), ∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
#oc(a1, y0) + #oc(a1, y) < #oc(a1, y -1l a), ∀ a0. (#oc(a0, a1 ') + #oc(a0, y0)) + #oc(a0, y) ≤ #oc(a0, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ (#oc(a0, a1 ') + #oc(a0, y0)) + #oc(a0, y) ⊦

Rule argument

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1; ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y); ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic a: ⊦ (x + y) + z = x + y + z; from specification polylist ⊦ a + x = a ' + x; (s) c: ⊦ m + n = n + m; from specification nat-basic ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ a ' + x ≠ []; (s)

Proof step 12 from proof delapp (previous step) (back to tree)

#oc(a1, y0) + #oc(a1, y) < #oc(a1, y -1l a), ∀ a0. (#oc(a0, a1 ') + #oc(a0, y0)) + #oc(a0, y) ≤ #oc(a0, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ (#oc(a0, a1 ') + #oc(a0, y0)) + #oc(a0, y) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x)
Proof step 14:
a1 = a, #oc(a1, y0) + #oc(a1, y) < #oc(a1, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ (#oc(a0, a1 ') + #oc(a0, y0)) + #oc(a0, y), ∀ a0. (#oc(a0, a1 ') + #oc(a0, y0)) + #oc(a0, y) ≤ #oc(a0, y -1l a) ⊦

Rule argument

spec: <toplevel>, name:del1-01, seq: ⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x), subst:[b, x, a] → [a, y, a1], <all paths>

Additional information

Used simplifier rules: ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: a, y, a1

Proof step 13 from proof delapp (previous step) (back to tree)

⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x)

Lemma del1-01

Proof step 14 from proof delapp (previous step) (back to tree)

a1 = a, #oc(a1, y0) + #oc(a1, y) < #oc(a1, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ (#oc(a0, a1 ') + #oc(a0, y0)) + #oc(a0, y), ∀ a0. (#oc(a0, a1 ') + #oc(a0, y0)) + #oc(a0, y) ≤ #oc(a0, y -1l a) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
#oc(a, y0) + #oc(a, y) < #oc(a, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ (#oc(a0, a ') + #oc(a0, y0)) + #oc(a0, y), ∀ a0. (#oc(a0, a ') + #oc(a0, y0)) + #oc(a0, y) ≤ #oc(a0, y -1l a) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 15 from proof delapp (previous step) (back to tree)

#oc(a, y0) + #oc(a, y) < #oc(a, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ (#oc(a0, a ') + #oc(a0, y0)) + #oc(a0, y), ∀ a0. (#oc(a0, a ') + #oc(a0, y0)) + #oc(a0, y) ≤ #oc(a0, y -1l a) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ a ∈ x → #oc(a, x -1l a) = #oc(a, x) -1

Rule argument

spec: <toplevel>, name:del1, seq: ⊦ a ∈ x → #oc(a, x -1l a) = #oc(a, x) -1, subst:[x, a] → [y, a], <all paths>

Additional information

Used simplifier rules: notin: ⊦ ¬ a ∈ x → #oc(a, x) = 0; in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0; ⊦ n -1 = n - 1; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < 1 ↔ m = 0; (s) n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ 0 + m = m; (s) ⊦ ¬ a ∈ x → x -1l a = x; (s) ⊦ ¬ n < 0; (s) Used terms: y, a

Proof step 16 from proof delapp (previous step) (back to tree)

⊦ a ∈ x → #oc(a, x -1l a) = #oc(a, x) -1

Lemma del1

Proof Tree for delapp (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.