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

x = a + y, (m0 + 1) + n < # x, ∀ n0, x0. m0 + n0 < # x0 → restn(m0, x0).first ' + firstn(n0, restn(m0, x0).rest) = firstn(n0, restn(m0, x0)) + x0[m0 + n0] ' ⊦ x = [], restn(m0, y).first ' + firstn(n, restn(m0, y).rest) = firstn(n, restn(m0, y)) + y[m0 + n] '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m0 + n < # y, ∀ n0, x0. m0 + n0 < # x0 → restn(m0, x0).first ' + firstn(n0, restn(m0, x0).rest) = firstn(n0, restn(m0, x0)) + x0[m0 + n0] ' ⊦ restn(m0, y).first ' + firstn(n, restn(m0, y).rest) = firstn(n, restn(m0, y)) + y[m0 + n] ', y = []

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list