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Proof step 8 from proof prefixn-shorter (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
prefixn(n, x) = prefixn(n, x0), ∀ n, x1. prefixn(n + 1, x) = prefixn(n + 1, x1) → prefixn(n, x) = prefixn(n, x1) ⊦ prefixn(n, a0 ' + x) = prefixn(n, a0 ' + x0)

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' + x ↔ a = b ∨ a ∈ x; (s)% ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ prefixn(n + 1, a ' + x) = a ' + prefixn(n, x); (s)% ⊦ a + x = a ' + x; (s)