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

n1 ≠ 0, n1 + m0 < # x, n0 < n1, ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).rest) ⊑ firstn(n1, restn(m0, x).rest) ⊦ x = [], n1 = 0, restn(m0, x).rest.first ' + firstn(n0, restn(m0, x).rest.rest) ⊑ restn(m0, x).rest.first ' + firstn(n1 -1, restn(m0, x).rest.rest)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n1 + m0 < # x, n0 < n1, ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).rest) ⊑ firstn(n1, restn(m0, x).rest) ⊦ x = [], n1 = 0, firstn(n0, restn(m0, x).rest.rest) ⊑ firstn(n1 - 1, restn(m0, x).rest.rest)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%prefix-rec-02: ⊦ b ' + x ⊑ b ' + y ↔ x ⊑ y; from specification list-last%pred-03: ⊦ n -1 = n - 1; from specification nat