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Proof step 6 from proof put-firstn-restn (previous step)

y = y0 + z ∧ # y0 = n, n ≤ # y ⊦ (a0 ' + y)[n, a] = firstn(n, a0 ' + y) + a ' + z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ (a0 ' + y0 + z)[# y0, a] = firstn(# y0, a0 ' + y0 + z) + a ' + z

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list