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Proof step 4 from proof firstn-butlast (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a0 ' + x0 = firstn(# x0 - # 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), a0 ∈ firstn(# x0 - # z, a0 ' + x0) ∨ a0 = a ∨ a0 ∈ z ⊦ firstn(1 + # x0 - # z, a0 ' + x0).butlast = firstn(# x0 - # z, a0 ' + x0)

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%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% ⊦ m + n - n + n1 = m - n1; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)