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Proof step 1 from proof last-ex (conclusion)

a ≠ x.last, a ∈ x ⊦ ∃ y, z. x = y + a ' + z ∧ z ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ x ⊦ a = x.last, ∃ y, z. x = y + a ' + z ∧ z ≠ []

Rule argument

Additional information