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Proof step 6 from proof sublist-part (previous step)

m0 ≠ 0, # x0 + 1 = n0 + m0, ∀ n0, m0. # x0 = n0 + m0 → firstn(m0, x0) + firstn(n0, restn(m0, x0)) = x0 ⊦ ((a ' + x0).first ' + firstn(m0 -1, (a ' + x0).rest)) + firstn(n0, restn(m0, a ' + x0)) = a ' + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# x0 + 1 = n0 + m0, ∀ n0, m0. # x0 = n0 + m0 → firstn(m0, x0) + firstn(n0, restn(m0, x0)) = x0 ⊦ m0 = 0, firstn(m0 - 1, x0) + firstn(n0, restn(m0, a ' + x0)) = x0

Rule argument

Additional information

Used simplifier rules: app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%pred-03: ⊦ n -1 = n - 1; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%c-08: ⊦ (a ' + x).first = a; from specification list