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Proof step 12 from proof lastn-restn (previous step)

m < #(a ' + x0), m + m0 ≤ # x0, ∀ m, n. m + n ≤ # x0 → lastn(m, restn(n, x0)) = lastn(m, x0) ⊦ lastn(m, restn(m0, x0)) = lastn(m, (a ' + x0).tail)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
m + m0 ≤ # x0, m ≤ # x0, ∀ m, n. m + n ≤ # x0 → lastn(m, restn(n, x0)) = lastn(m, x0) ⊦ lastn(m, restn(m0, x0)) = lastn(m, x0)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)