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Proof step 3 from proof firstn-butlast (previous step)

x = firstn(# x - 1 + # z, x) + a ' + z ⊦ firstn(# x - # z, x).butlast = firstn(# x - 1 + # z, x)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
a0 + x0 = firstn(#(a0 + x0) - 1 + # z, a0 + x0) + a ' + z, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(# x0 - # z, x0).butlast = firstn(# x0 - 1 + # z, x0) ⊦ firstn(#(a0 + x0) - # z, a0 + x0).butlast = firstn(#(a0 + x0) - 1 + # z, a0 + x0)

Rule argument

indvar: x

Additional information

Used simplifier rules: firstN-empty: ⊦ firstn(n, x) = [] ↔ n = 0;%list freely generated by [], +;;% ⊦ # [] = 0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ a ' + x ≠ []; (s)