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Proof step 6 from proof restn-nonempty-2 (previous step)

0 < n, restn(n -1, (a ' + x0).rest) = [], n ≤ # x0, (∃ n. restn(n, x0) = [] ∧ n < # x0) → x0 = [] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
restn(n - 1, x0) = [], n ≤ # x0 ⊦ n = 0, x0 = [], ∃ n. restn(n, x0) = [] ∧ n < # x0

Rule argument

Additional information

Used simplifier rules: lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%size-01: ⊦ # [] = 0; from specification list-data%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%pred-03: ⊦ n -1 = n - 1; from specification nat