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Proof step 5 from proof restn-firstn-better (previous step)

n0 + m ≤ # x, m ≤ # x ⊦ restn(m, firstn(n0 + m, x)) = firstn(n0, restn(m, x))

User applied rule structural induction

And obtained the following new goals

Proof step 6:
m0 +1 ≤ # x, n0 + m0 +1 ≤ # x, ∀ n0, x. n0 + m0 ≤ # x ∧ m0 ≤ # x → restn(m0, firstn(n0 + m0, x)) = firstn(n0, restn(m0, x)) ⊦ restn(m0 +1, firstn(n0 + m0 +1, x)) = firstn(n0, restn(m0 +1, x))

Rule argument

indvar: m

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;%nat freely generated by 0, +1;;% ⊦ 0 + m = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic