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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. ¬ # y + # z < n ∧ (y + a ' + z)[n] = a

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list