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Proof step 12 from proof prefixn-shorter (previous step)

n = m + 1, 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(m, x) = prefixn(m, x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
prefixn(m + 1, x) = prefixn(m + 1, x0), ∀ n, x1. prefixn(n + 1, x) = prefixn(n + 1, x1) → prefixn(n, x) = prefixn(n, x1) ⊦ prefixn(m, x) = prefixn(m, x0)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)