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Proof step 9 from proof sublist-prefix (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 10:
(m + 1) + m0 < # x, n0 ≤ m, m + m0 < # x ∧ n0 ≤ m → x = [] ∨ firstn(n0, restn(m0, x).rest) ⊑ firstn(m, restn(m0, x).rest), ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).rest) ⊑ firstn(n1, restn(m0, x).rest) ⊦ x = [], firstn(n0, restn(m0, x).rest.rest) ⊑ firstn(m, restn(m0, x).rest.rest)

Rule argument

pos: left 3 terms: m, m0, x abort: F discard: F computed: F

Additional information

Used terms: m, m0, x