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Proof step 6 from proof app (previous step)

a ' + x0 ≠ [] ∧ (a ' + x0).rest ≠ [], ordered≤((a ' + x0) + y), (∃ y. ordered≤(x0 + y)) → ordered≤(x0) ⊦ ¬ (a ' + x0).rest.first < (a ' + x0).first ∧ ordered≤((a ' + x0).rest)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
ordered≤((a ' + x0) + y), (∃ y. ordered≤(x0 + y)) → ordered≤(x0) ⊦ x0 = [], ¬ x0.first < a ∧ ordered≤(x0)

Rule argument

Additional information

Used simplifier rules: c-08: ⊦ (a ' + x).first = a; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list