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Proof step 1 from proof shortest-frag (conclusion) (back to tree)

frag*(IOAut(start, step, ext))(sl, al) ⊦ ∃ sl0, al0. sl0.head = sl.head ∧ sl0.last = sl.last ∧ frag*(IOAut(start, step, ext))(sl0, al0) ∧ (∀ sl1, al1. sl1.head = sl.head ∧ sl1.last = sl.last ∧ frag*(IOAut(start, step, ext))(sl1, al1) → # sl0 ≤ # sl1)

User applied rule induction

And obtained the following new goals

Rule argument

indvar: # sl wfpred: <

Additional information

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

Proof step 2 from proof shortest-frag (previous step) (back to tree)

frag*(IOAut(start, step, ext))(sl, al), ∀ sl0, start, step, ext, al. # sl0 < # sl → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦ ∃ sl0, al0. sl0.head = sl.head ∧ sl0.last = sl.last ∧ frag*(IOAut(start, step, ext))(sl0, al0) ∧ (∀ sl1, al1. sl1.head = sl.head ∧ sl1.last = sl.last ∧ frag*(IOAut(start, step, ext))(sl1, al1) → # sl0 ≤ # sl1)

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: sl, al abort: F discard: F computed: T

Additional information

Used terms: sl, al

Proof step 3 from proof shortest-frag (previous step) (back to tree)

frag*(IOAut(start, step, ext))(sl, al), ∀ sl0, start, step, ext, al. # sl0 < # sl → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦ ∃ sl0, al0. sl0.head = sl.head ∧ sl0.last = sl.last ∧ frag*(IOAut(start, step, ext))(sl0, al0) ∧ (∀ sl1, al1. sl1.head = sl.head ∧ sl1.last = sl.last ∧ frag*(IOAut(start, step, ext))(sl1, al1) → # sl0 ≤ # sl1), sl.head = sl.head ∧ sl.last = sl.last ∧ frag*(IOAut(start, step, ext))(sl, al) ∧ (∀ sl1, al1. sl1.head = sl.head ∧ sl1.last = sl.last ∧ frag*(IOAut(start, step, ext))(sl1, al1) → # sl ≤ # sl1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≤ n ↔ ¬ n < m; (ss)

Proof step 4 from proof shortest-frag (previous step) (back to tree)

sl1.head = sl.head, sl1.last = sl.last, sl ≠ [], # sl1 < # sl, frag*(IOAut(start, step, ext))(sl1, al1), frag*(IOAut(start, step, ext))(sl, al), ∀ sl0, start, step, ext, al. # sl0 < # sl → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦ ∃ sl0, al0. sl0.head = sl.head ∧ sl0.last = sl.last ∧ frag*(IOAut(start, step, ext))(sl0, al0) ∧ (∀ sl1, al1. sl1.head = sl.head ∧ sl1.last = sl.last ∧ frag*(IOAut(start, step, ext))(sl1, al1) → # sl0 ≤ # sl1)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: list-data, name:elim-add, seq: ⊦ x ≠ [] → (head = x.head ∧ tail = x.tail ↔ x = head + tail), subst:[x, head, tail] → [sl, s, sl3]; ['a] → ['s]

Additional information

Used simplifier rules: ⊦ x ≠ [] → (head = x.head ∧ tail = x.tail ↔ x = head + tail); (elim) Used terms: sl, head, tail

Proof step 5 from proof shortest-frag (previous step) (back to tree)

sl1.head = s, sl1.last = sl.last, sl = s + sl3, sl ≠ [], # sl1 < # sl, frag*(IOAut(start, step, ext))(sl, al), frag*(IOAut(start, step, ext))(sl1, al1), ∀ sl0, start, step, ext, al. # sl0 < # sl → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦ ∃ sl0, al0. sl0.head = s ∧ sl0.last = sl.last ∧ frag*(IOAut(start, step, ext))(sl0, al0) ∧ (∀ sl1, al1. sl1.head = s ∧ sl1.last = sl.last ∧ frag*(IOAut(start, step, ext))(sl1, al1) → # sl0 ≤ # sl1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s) ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ; (s) ⊦ a ' + x ≠ []; (s)

Proof step 6 from proof shortest-frag (previous step) (back to tree)

sl1.last = (sl1.head ' + sl3).last, frag*(IOAut(start, step, ext))(sl1, al1), frag*(IOAut(start, step, ext))(sl1.head ' + sl3, al), # sl1 ≤ # sl3, ∀ sl0, start, step, ext, al. # sl0 < #(sl1.head + sl3) → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦ ∃ sl0, al0. sl0.head = sl1.head ∧ sl0.last = sl1.last ∧ frag*(IOAut(start, step, ext))(sl0, al0) ∧ (∀ sl, al2. sl.head = sl1.head ∧ sl.last = sl1.last ∧ frag*(IOAut(start, step, ext))(sl, al2) → # sl0 ≤ # sl)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: list-data, name:elim-list-c, seq:x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y, subst:[x, a, y] → [sl1, s, sl4]; ['a] → ['s]

Additional information

Used simplifier rules: not-frag-empty: ⊦ ¬ frag(IOAut(start, step, ext))([], α); ⊦ # [] = 0; (s) ⊦ 0 ≤ m; (s) ⊦ ⌜[]⌝ = []; (s) x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: sl1, s, y

Proof step 7 from proof shortest-frag (previous step) (back to tree)

sl1.last = (s ' + sl3).last, sl1 = s + sl4, sl1 ≠ [], frag*(IOAut(start, step, ext))(s ' + sl3, al), frag*(IOAut(start, step, ext))(sl1, al1), # sl1 ≤ # sl3, ∀ sl0, start, step, ext, al. # sl0 < #(s + sl3) → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦ ∃ sl0, al0. sl0.head = s ∧ sl0.last = sl1.last ∧ frag*(IOAut(start, step, ext))(sl0, al0) ∧ (∀ sl, al2. sl.head = s ∧ sl.last = sl1.last ∧ frag*(IOAut(start, step, ext))(sl, al2) → # sl0 ≤ # sl)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ x ≠ [] → (a ' + x).last = x.last; (s) ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ a + x = a ' + x; (s) ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s) ⊦ a ' + x ≠ []; (s)

Proof step 8 from proof shortest-frag (previous step) (back to tree)

(s ' + sl4).last = sl3.last, sl3 ≠ [], # sl4 < # sl3, frag*(IOAut(start, step, ext))(s ' + sl4, al1), frag*(IOAut(start, step, ext))(s ' + sl3, al), ∀ sl0, start, step, ext, al. # sl0 < #(s + sl3) → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦ ∃ sl0, al0. sl0.head = s ∧ sl0.last = (sl0.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl0, al0) ∧ (∀ sl, al2. sl.head = sl0.head ∧ sl.last = (sl0.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl, al2) → # sl0 ≤ # sl)

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [sl0, start, step, ext, al] → [s + sl4, start, step, ext, al1]

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s) ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ #(a ' + x) = (# x)+1; (s) ⊦ m +1 = m + 1; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → (n ≤ m ↔ false); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: s + sl4, start, step, ext, al1

Proof step 9 from proof shortest-frag (previous step) (back to tree)

(s ' + sl4).last = sl3.last, sl3 ≠ [], # sl4 < # sl3, frag*(IOAut(start, step, ext))(s ' + sl3, al), frag*(IOAut(start, step, ext))(s ' + sl4, al1), ∃ sl1, al0. sl1.head = (s + sl4).head ∧ sl1.last = (s + sl4).last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = (s + sl4).head ∧ sl2.last = (s + sl4).last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2), ∀ sl0, start, step, ext, al. # sl0 < #(s + sl3) → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦ ∃ sl0, al0. sl0.head = s ∧ sl0.last = (sl0.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl0, al0) ∧ (∀ sl, al2. sl.head = sl0.head ∧ sl.last = (sl0.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl, al2) → # sl0 ≤ # sl)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ (a ' + x).head = a; (s)

Proof step 10 from proof shortest-frag (previous step) (back to tree)

sl3.last = (sl1.head ' + sl4).last, sl1.last = sl3.last, sl3 ≠ [], # sl4 < # sl3, frag*(IOAut(start, step, ext))(sl1.head ' + sl3, al), frag*(IOAut(start, step, ext))(sl1.head ' + sl4, al1), frag*(IOAut(start, step, ext))(sl1, al0), ∀ sl2, al2. sl2.head = sl1.head ∧ sl2.last = (sl2.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl2, al2) → # sl1 ≤ # sl2, ∀ sl0, start, step, ext, al. # sl0 < #(sl1.head + sl3) → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦ ∃ sl0, al0. sl0.head = sl1.head ∧ sl0.last = (sl0.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl0, al0) ∧ (∀ sl, al2. sl.head = sl0.head ∧ sl.last = (sl0.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl, al2) → # sl0 ≤ # sl)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: list-data, name:elim-list-c, seq:x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y, subst:[x, a, y] → [sl1, s, sl5]; ['a] → ['s]

Additional information

Used simplifier rules: not-frag-empty: ⊦ ¬ frag(IOAut(start, step, ext))([], α); ⊦ ⌜[]⌝ = []; (s) x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: sl1, s, y

Proof step 11 from proof shortest-frag (previous step) (back to tree)

sl1.last = sl3.last, sl3.last = (s ' + sl4).last, sl1 = s + sl5, sl1 ≠ [], sl3 ≠ [], # sl4 < # sl3, frag*(IOAut(start, step, ext))(sl1, al0), frag*(IOAut(start, step, ext))(s ' + sl4, al1), frag*(IOAut(start, step, ext))(s ' + sl3, al), ∀ sl2, al2. sl2.head = s ∧ sl2.last = (sl2.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl2, al2) → # sl1 ≤ # sl2, ∀ sl0, start, step, ext, al. # sl0 < #(s + sl3) → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦ ∃ sl0, al0. sl0.head = s ∧ sl0.last = (sl0.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl0, al0) ∧ (∀ sl, al2. sl.head = sl0.head ∧ sl.last = (sl0.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl, al2) → # sl0 ≤ # sl)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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 + 1 ≤ n ↔ m < n; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s) ⊦ a + x = a ' + x; (s) ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ; (s) ⊦ a ' + x ≠ []; (s)

Proof step 12 from proof shortest-frag (previous step) (back to tree)

sl3.last = (s ' + sl4).last, (s ' + sl5).last = sl3.last, sl3 ≠ [], # sl4 < # sl3, frag*(IOAut(start, step, ext))(s ' + sl3, al), frag*(IOAut(start, step, ext))(s ' + sl4, al1), frag*(IOAut(start, step, ext))(s ' + sl5, al0), ∀ sl2, al2. sl2.head = s ∧ sl2.last = (sl2.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl2, al2) → # sl5 < # sl2, ∀ sl0, start, step, ext, al. # sl0 < #(s + sl3) → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦ ∃ sl0, al0. sl0.head = s ∧ sl0.last = (sl0.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl0, al0) ∧ (∀ sl, al2. sl.head = sl0.head ∧ sl.last = (sl0.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl, al2) → # sl0 ≤ # sl)

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: s + sl5, al0 abort: F discard: T computed: F

Additional information

Used terms: s + sl5, al0

Proof step 13 from proof shortest-frag (previous step) (back to tree)

(s ' + sl5).last = sl3.last, sl3.last = (s ' + sl4).last, sl3 ≠ [], # sl4 < # sl3, frag*(IOAut(start, step, ext))(s ' + sl5, al0), frag*(IOAut(start, step, ext))(s ' + sl4, al1), frag*(IOAut(start, step, ext))(s ' + sl3, al), ∀ sl2, al2. sl2.head = s ∧ sl2.last = (sl2.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl2, al2) → # sl5 < # sl2, ∀ sl0, start, step, ext, al. # sl0 < #(s + sl3) → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦ (s + sl5).head = s ∧ (s + sl5).last = ((s + sl5).head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(s + sl5, al0) ∧ (∀ sl, al2. sl.head = (s + sl5).head ∧ sl.last = ((s + sl5).head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl, al2) → #(s + sl5) ≤ # sl)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≤ n ↔ m < n; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s) ⊦ a + x = a ' + x; (s) ⊦ (a ' + x).head = a; (s) ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ; (s)

Proof step 14 from proof shortest-frag (previous step) (back to tree)

sl.last = sl3.last, sl3.last = (sl.head ' + sl4).last, (sl.head ' + sl5).last = sl3.last, sl3 ≠ [], # sl4 < # sl3, frag*(IOAut(start, step, ext))(sl, al2), frag*(IOAut(start, step, ext))(sl.head ' + sl3, al), frag*(IOAut(start, step, ext))(sl.head ' + sl4, al1), frag*(IOAut(start, step, ext))(sl.head ' + sl5, al0), # sl ≤ # sl5, ∀ sl2, al2. sl2.head = sl.head ∧ sl2.last = (sl2.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl2, al2) → # sl5 < # sl2, ∀ sl0, start, step, ext, al. # sl0 < #(sl.head + sl3) → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: list-data, name:elim-list-c, seq:x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y, subst:[x, a, y] → [sl, s, sl6]; ['a] → ['s]

Additional information

Used simplifier rules: not-frag-empty: ⊦ ¬ frag(IOAut(start, step, ext))([], α); ⊦ # [] = 0; (s) ⊦ 0 ≤ m; (s) ⊦ ⌜[]⌝ = []; (s) x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: sl, s, y

Proof step 15 from proof shortest-frag (previous step) (back to tree)

(s ' + sl5).last = sl3.last, sl3.last = (s ' + sl4).last, sl.last = sl3.last, sl = s + sl6, sl ≠ [], sl3 ≠ [], # sl4 < # sl3, frag*(IOAut(start, step, ext))(s ' + sl5, al0), frag*(IOAut(start, step, ext))(s ' + sl4, al1), frag*(IOAut(start, step, ext))(s ' + sl3, al), frag*(IOAut(start, step, ext))(sl, al2), # sl ≤ # sl5, ∀ sl2, al2. sl2.head = s ∧ sl2.last = (sl2.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl2, al2) → # sl5 < # sl2, ∀ sl0, start, step, ext, al. # sl0 < #(s + sl3) → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ x ≠ [] → (a ' + x).last = x.last; (s) ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ; (s) ⊦ a + x = a ' + x; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ m +1 = m + 1; (s) ⊦ #(a ' + x) = (# x)+1; (s) ⊦ a ' + x ≠ []; (s)

Proof step 16 from proof shortest-frag (previous step) (back to tree)

sl3.last = (s ' + sl4).last, (s ' + sl6).last = sl3.last, sl5.last = sl3.last, sl3 ≠ [], sl5 ≠ [], # sl6 < # sl5, # sl4 < # sl3, frag*(IOAut(start, step, ext))(s ' + sl6, al2), frag*(IOAut(start, step, ext))(s ' + sl5, al0), frag*(IOAut(start, step, ext))(s ' + sl4, al1), frag*(IOAut(start, step, ext))(s ' + sl3, al), ∀ sl2, al2. sl2.head = s ∧ sl2.last = (sl2.head ' + sl4).last ∧ frag*(IOAut(start, step, ext))(sl2, al2) → # sl5 < # sl2, ∀ sl0, start, step, ext, al. # sl0 < #(s + sl3) → frag*(IOAut(start, step, ext))(sl0, al) → (∃ sl1, al0. sl1.head = sl0.head ∧ sl1.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl1, al0) ∧ (∀ sl2, al1. sl2.head = sl0.head ∧ sl2.last = sl0.last ∧ frag*(IOAut(start, step, ext))(sl2, al1) → # sl1 ≤ # sl2)) ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 12 terms: s + sl6, al2 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ (a ' + x).head = a; (s) ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ; (s) ⊦ a + x = a ' + x; (s) ⊦ #(a ' + x) = (# x)+1; (s) ⊦ m +1 = m + 1; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m < n + 1 ↔ ¬ 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: s + sl6, al2

Proof Tree for shortest-frag (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.

the proof tree