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Proof step 10 from proof ordered-append (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 11:
ordered≤(a ' + x1), ordered≤(a0 ' + y), ordered≤(a0 ' + y) ∧ (a0 ' + y ≠ [] → ¬ (a0 ' + y).head < x1.last) → ordered≤(x1 + a0 ' + y), ∀ x. ordered≤(x) ∧ (x ≠ [] → ¬ x.head < x1.last) → ordered≤(x1 + x) ⊦ x1 = [], a0 < x1.last, ordered≤(a ' + x1 + a0 ' + y)

Rule argument

pos: left 3 terms: a0 ' + y abort: F discard: F computed: T

Additional information

Used terms: a0 ' + y