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Proof step 19 from proof ordered-prefix (previous step)

y ⊑ y0, ordered≤(a0 ' + y0) ⊦ a0 < a, ordered≤(a0 ' + y), ∃ x. a0 ' + y ⊑ x ∧ ordered≤(x)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: a0 ' + y0

Additional information

Used simplifier rules: prefix-rec-03: ⊦ a ' + x ⊑ b ' + y ↔ a = b ∧ x ⊑ y; from specification list-last%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a0 ' + y0