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Proof step 14 from proof firstn-butlast (previous step)

# x0 = n0 + # z, a0 ' + x0 = (a0 ' + firstn(n0 -1, x0)) + a ' + z, # z ≤ # x0, x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(n0, x0).butlast = firstn(# x0 - 1 + # z, x0), a0 ∈ a0 ' + firstn(n0 -1, x0) ∨ a0 = a ∨ a0 ∈ z ⊦ (a0 ' + firstn(n0, x0)).butlast = a0 ' + firstn(n0 -1, x0), n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# x0 = n0 + # z, x0 = firstn(n0 - 1, x0) + a ' + z, # z ≤ # x0, x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(n0, x0).butlast = firstn(# x0 - 1 + # z, x0) ⊦ (a0 ' + firstn(n0, x0)).butlast = a0 ' + firstn(n0 - 1, x0), n0 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ x + y = x + z ↔ y = z; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ n -1 = n - 1; (s)