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Proof step 6 from proof expand-cons (previous step)

x0 = [] ∨ ⌜x0⌝ = ⌜a0 '⌝ + σ ⊦ ∃ a1, σ0. ⌜a '⌝ + ⌜x0⌝ = ⌜a1 '⌝ + σ0

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: a, ⌜x0⌝

Additional information

Used terms: a, ⌜x0⌝