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Proof step 1 from proof get-from-01 (conclusion)

m + m0 < # x ⊦ (x from m)[m0] = x[m + m0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m + m0 < # x ⊦ (x from m)[m0] = x[m + m0], x = []

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