[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 4 from proof put-firstn-restn (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
⊦ n ≤ # x → (y = firstn(n, x) ∧ z = restn(n, x) ↔ x = y + z ∧ # y = n)
Proof step 6:
y = y0 + z ∧ # y0 = n, n ≤ # y ⊦ (a0 ' + y)[n, a] = firstn(n, a0 ' + 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] → [y, n, y0, z] flags: rewrite, current

Additional information

Used terms: y, n, y0, z