[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 17 from proof refine-add (previous step)

xs_len + 1 = h, H0 = (H1 ++ r)[r, (H1 ++ r)[r].val:= a][r, (H1 ++ r)[r, (H1 ++ r)[r].val:= a][r].nxt:= rxs], h = h0, xs_len = # xs, ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n)))(H), ((del-list(to_delete, n) * ls(rys, ys) * ls(rxs, xs)))(H), # xs < h0, n ≤ # xs + # ys, n + # xs ≤ h0, n < 2 → to_delete0 = [] ∧ ((ls(rys, ys) * ls(rxs, xs)))(H1), 2 ≤ n → ((del-list(to_delete0, n - 2) * ls(rys, ys) * ls(rxs, xs)))(H1), ys = [] ∨ # ys = h0 ⊦ ⟪if rys ≠ null then to_delete0 := rys + to_delete0; rys := r; xs_len := 0; rxs := null⟫ ⟨xs_len0 := # xs; caterpillar_add#(a, h0; xs, xs_len0, ys)⟩ ∃ n. ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete0, n)))(H0) ∧ n ≤ # xs + # ys ∧ n + # xs ≤ h0 ∧ xs_len = xs_len0, r = null, r ∈ H1

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

pos: right 1

Additional information