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Proof step 10 from proof firstn-butlast (previous step)

# x0 = n0 + # z, # x0 = n1 + # z, a0 ' + x0 = firstn(n1, a0 ' + x0) + a ' + z, # z ≤ # x0, x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(n0, x0).butlast = firstn(# x0 - 1 + # z, x0), a0 ∈ firstn(n1, a0 ' + x0) ∨ a0 = a ∨ a0 ∈ z ⊦ (a0 ' + firstn(n1, x0)).butlast = firstn(n1, a0 ' + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# x0 = n0 + # z, a0 ' + x0 = firstn(n0, a0 ' + 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, a0 ' + x0) ∨ a0 = a ∨ a0 ∈ z ⊦ (a0 ' + firstn(n0, x0)).butlast = firstn(n0, a0 ' + x0)

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2