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Proof step 20 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) ⊦ (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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
x0 + a ' = a ' + y, ordered≤(a ' + y), a ∈ y, ∀ a, b. ordered≤(x0 + a ') → (ordered≤(x0 + a ' + b ') ↔ ¬ b < a) ⊦ b < a, a < a0, b < a0, ordered≤(x0 + a ' + b ')

Rule argument

Additional information

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