[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 207 from proof hasSharing-put-index-refval (previous step)

(st[r -- i ', refval(r2)][r2 -- (sks0 + i ' ' + sks)]) = refval(r1), st[r2 -- sks1] = refval(r1), st[r -- _type] .type = mkarraytype(ty), st[r2 -- sks0] = refval(r), 0 < st[r -- _length] .intval, fits(r2, sks0 + i ' ' + sks, st), 0 ≤ i, i < st[r -- _length] .intval, fits(r2, sks1, st), is_referencevalue(st[r -- i ', refval(r2)][r -- (i ' ' + sks1)]), validref(str, r2, ty, st), disjoint(str, r ' + r2 ', st), ∀ sks2. sks2 + i ' ' ⊑ sks0 → st[r2 -- sks2] .refval ≠ r ⊦ sks1 = sks0 + i ' ' + sks, r2 = jvmref, r1 = jvmref, r = r2, r = jvmref, hasSharing(r, st), hasSharing(r2, st), junkref(str, r, st), ∃ sks2, sks3. sks0 + i ' ' + sks = sks2 + i ' ' + sks3 ∧ st[r2 -- sks2] .refval = r ∧ (∀ sks. sks + i ' ' ⊑ sks2 → st[r2 -- sks] .refval ≠ st[r2 -- sks2] .refval), ∃ sks, sks2. sks1 = sks + i ' ' + sks2 ∧ st[r2 -- sks] .refval = r ∧ (∀ sks2. sks2 + i ' ' ⊑ sks → st[r2 -- sks2] .refval ≠ st[r2 -- sks] .refval)

User applied rule exists right

And obtained the following new goals

Proof step 206:
(st[r -- i ', refval(r2)][r2 -- (sks0 + i ' ' + sks)]) = refval(r1), st[r2 -- sks1] = refval(r1), st[r -- _type] .type = mkarraytype(ty), st[r2 -- sks0] = refval(r), 0 < st[r -- _length] .intval, fits(r2, sks0 + i ' ' + sks, st), 0 ≤ i, i < st[r -- _length] .intval, fits(r2, sks1, st), is_referencevalue(st[r -- i ', refval(r2)][r -- (i ' ' + sks1)]), validref(str, r2, ty, st), disjoint(str, r ' + r2 ', st), ∀ sks2. sks2 + i ' ' ⊑ sks0 → st[r2 -- sks2] .refval ≠ r ⊦ sks1 = sks0 + i ' ' + sks, r2 = jvmref, r1 = jvmref, r = r2, r = jvmref, hasSharing(r, st), hasSharing(r2, st), junkref(str, r, st), sks0 + i ' ' + sks = sks0 + i ' ' + sks ∧ st[r2 -- sks0] .refval = r ∧ (∀ sks. sks + i ' ' ⊑ sks0 → st[r2 -- sks] .refval ≠ st[r2 -- sks0] .refval), ∃ sks, sks2. sks1 = sks + i ' ' + sks2 ∧ st[r2 -- sks] .refval = r ∧ (∀ sks2. sks2 + i ' ' ⊑ sks → st[r2 -- sks2] .refval ≠ st[r2 -- sks] .refval)

Rule argument

[EXRARG : pos: right 9, [extquanttermlist: (sks0, sks) abort:F , discard:T , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sks0, sks)]]