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Proof step 4 from proof in-get (previous step)

∃ y, z. x = y + (a + z) ⊦ ∃ n. n < # x ∧ x[n] = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ ∃ n. n < # y + 1 + # z ∧ (y + a ' + z)[n] = a

Rule argument

Additional information

Used simplifier rules: ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)