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Proof step 6 from proof prefixn-shorter (previous step)

prefixn(n + 1, a + x) = 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 7:
a ' + prefixn(n, x) = prefixn(n + 1, x1), 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)

Rule argument

Additional information

Used simplifier rules: in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%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)