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Proof step 9 from proof ordered-append (previous step)

ordered≤(a0 ' + y), ordered≤(a ' + x1), ordered≤(a0 ' + y) ∧ (a0 ' + y ≠ [] → ¬ (a0 ' + y).first < x1.last) → ordered≤(x1 + a0 ' + y), ∀ x. ordered≤(x) ∧ (x ≠ [] → ¬ x.first < x1.last) → ordered≤(x1 + x) ⊦ x1 = [], ordered≤(a ' + x1 + a0 ' + y), a0 < x1.last

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
ordered≤(a ' + x1), ordered≤(x1 + a0 ' + y), ∀ x. ordered≤(x) ∧ (x ≠ [] → ¬ x.first < x1.last) → ordered≤(x1 + x) ⊦ x1 = [], ordered≤(a ' + x1 + a0 ' + y), a0 < x1.last

Rule argument

Additional information

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