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Proof step 10 from proof prefixn-shorter (previous step)

prefixn(n, x) = prefixn(n, x0), ∀ n, x1. prefixn(n + 1, x) = prefixn(n + 1, x1) → prefixn(n, x) = prefixn(n, x1) ⊦ n = 0, prefixn(n, a0 ' + x) = prefixn(n, a0 ' + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
prefixn(n, x) = prefixn(n, x0), ∀ n, x1. prefixn(n + 1, x) = prefixn(n + 1, x1) → prefixn(n, x) = prefixn(n, x1) ⊦ n = 0, prefixn(n - 1, x) = prefixn(n - 1, x0)

Rule argument

Additional information

Used simplifier rules: prefixn-add: n ≠ 0 ⊦ prefixn(n, a ' + x0) = prefixn(n, a ' + x1) ↔ prefixn(n - 1, x0) = prefixn(n - 1, x1);%a: ⊦ (x + y) + z = x + y + z; from specification list