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Proof step 7 from proof put-firstn-restn (previous step)

⊦ (a0 ' + y0 + z)[# y0, a] = firstn(# y0, a0 ' + y0 + z) + a ' + z

User applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
⊦ n ≤ # x → (y = firstn(n, x) ∧ z = restn(n, x) ↔ x = y + z ∧ # y = n)
Proof step 9:
a0 ' + y0 + z = y + z0 ∧ # y = # y0 ∧ # y0 ≤ #(a0 ' + y0 + z) ⊦ (a0 ' + y0 + z)[# y0, a] = y + a ' + z

Rule argument

spec: list-del inst: name: elim-firstn-restn seq: ⊦ n ≤ # x → (y = firstn(n, x) ∧ z = restn(n, x) ↔ x = y + z ∧ # y = n) subst: [x, n, y, z] → [a0 ' + y0 + z, # y0, y, z0] flags: current, []

Additional information

Used simplifier rules: len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: a0 ' + y0 + z, # y0, y, z0