[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 30 from proof multi-length-first (previous step)

# refs + 1 = i→n(countrefs(i ' + ints0)), posints(ints0), 0 < n→i(# acc) + i, 0 < i, ∀ st, ty, r, acc, ints, i, refs0. # refs0 < # refs → ( true → ( # refs0 + 1 = i→n(countrefs(i ' + ints .rest)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) ∧ ¬ 0 ∈ ints .butlast ∧ i ≤ ints .first ∧ n→i(# acc) + i = ints .first → (addmultiarrays(r ' + refs0, i, acc, ty, ints, st)[r -- _length]) = intval(ints .first))) ⊦ ints0 = [], 0 = n→i(# acc) + i, i = 0, (addmultiarrays (firstn(1 + # refs - i→n(countrefs(ints0)), r ' + refs), i - 1, lastn(i→n(countrefs(ints0)), r ' + refs) .first ' + acc, ty, (n→i(# acc) + i) ' + ints0, addmultiarrays(lastn(i→n(countrefs(ints0)), r ' + refs), ints0 .first, [], ty, ints0, st)) [r -- _length ] ) = intval(n→i(# acc) + i), 0 ∈ ints0 .butlast

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

[speclemmaarg: ints ≠ [] ,i = ints .first ∧ ints0 = ints .rest ↔ ints = i + ints0 ,[ints, i, ints0] → [ints0, i0, ints1] ,T ,list-data ,intlista ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ints ≠ [] ⊦ i = ints .first ∧ ints0 = ints .rest ↔ ints = i + ints0])], [EXTRATERMS : (ints0, i0, ints1)]]