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Proof step 18 from proof ole-last (previous step)

x0 + a ' = a ' + y, ordered≤(a ' + y), a ∈ y, ∀ a, b. ordered≤(x0 + a ') → (ordered≤(x0 + a ' + b ') ↔ ¬ b < a) ⊦ b < a0, a < a0, b < a, ordered≤((a0 ' + x0) + a ' + b ')

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ x ≠ [] ∧ x.tail ≠ [] → (ordered≤(x) ↔ ¬ x.tail.head < x.head ∧ ordered≤(x.tail))
Proof step 20:
x0 + a ' = a ' + y, ordered≤(a ' + y), a ∈ y, ∀ a, b. ordered≤(x0 + a ') → (ordered≤(x0 + a ' + b ') ↔ ¬ b < a) ⊦ (a0 ' + x0) + a ' + b ' = [], ((a0 ' + x0) + a ' + b ').tail = [], b < a, a < a0, b < a0, ¬ ((a0 ' + x0) + a ' + b ').tail.head < ((a0 ' + x0) + a ' + b ').head ∧ ordered≤(((a0 ' + x0) + a ' + b ').tail)

Rule argument

spec: <toplevel> name: ole-rec seq: ⊦ x ≠ [] ∧ x.tail ≠ [] → (ordered≤(x) ↔ ¬ x.tail.head < x.head ∧ ordered≤(x.tail)) subst: [x] → [(a0 ' + x0) + a ' + b '] flags: <all paths>

Additional information

Used simplifier rules: ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)% ⊦ x + a ' ≠ []; (s) Used terms: (a0 ' + x0) + a ' + b '