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Proof step 19 from proof ole-last (previous step)

(a0 ' + x0) + a ' + b ' ≠ [] ∧ ((a0 ' + x0) + a ' + b ').rest ≠ [], x0 + a ' = a ' + y, ordered≤(a ' + y), a ∈ y, ∀ a, b. ordered≤(x0 + a ') → (ordered≤(x0 + a ' + b ') ↔ ¬ b < a) ⊦ b < a0, ¬ ((a0 ' + x0) + a ' + b ').rest.first < ((a0 ' + x0) + a ' + b ').first ∧ ordered≤(((a0 ' + x0) + a ' + b ').rest), a < a0, b < a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
x0 + a ' = a ' + y, a ∈ y, ordered≤(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: c-08: ⊦ (a ' + x).first = a; from specification list%first-02: ⊦ (x + a ' + y).first = (x + a ').first; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%e-02: ⊦ x + a ' ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list