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

(st[r -- i ', refval(r2)][r2 -- sks3]) = refval(r1), st[r -- _type] .type = mkarraytype(ty), (st[r -- i ', refval(r2)][r2 -- ((sks1 + i ' ' + sks4) + i ' ' + sks0)]) = refval(r1), st[r2 -- (sks1 + i ' ' + sks4)] = refval(r), (st[r -- i ', refval(r2)][r2 -- (sks1 + i ' ' + sks4)]) = refval(r0), st[r2 -- sks1] = refval(r), 0 < st[r -- _length] .intval, is_referencevalue(st[r -- i ', refval(r2)][r -- (i ' ' + (sks1 + i ' ' + sks4) + i ' ' + sks0)]), fits(r2, sks1 + i ' ' + sks4, st[r -- i ', refval(r2)]), fits(r0, i ' ' + sks0, st[r -- i ', refval(r2)]), i < st[r -- _length] .intval, 0 ≤ i, fits(r2, sks3, st[r -- i ', refval(r2)]), disjoint(str, r ' + r2 ', st), validref(str, r2, ty, st) ⊦ r = jvmref, r = r2, r1 = jvmref, r2 = jvmref, sks3 = [], (sks1 + i ' ' + sks4) + i ' ' + sks0 = sks3, junkref(str, r, st), hasSharing(r2, st), hasSharing(r, st), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks1 + i ' ' + sks4 → st[r2 -- sks2] .refval ≠ r), ∃ sks2. ¬ (sks2 + i ' ' ⊑ sks1 → st[r2 -- sks2] .refval ≠ r)

User applied rule exists right

And closed the goal!

Rule argument

[EXRARG : pos: right 10, [extquanttermlist: (sks1)abort:F, discard:T, computed:F]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPNAMEDSEQ : list-last, storekeys, app-01-02, ⊦ sks + sks0 ⊑ sks + sks1 ↔ sks0 ⊑ sks1], [CSIMPNAMEDSEQ : list-last, storekeys, prefix-one, ⊦ sk ' ⊑ sk0 ' + sks ↔ sk = sk0], [CSIMPNAMEDSEQ : javavalue, , sel-09, ⊦ refval(r) .refval = r])], [EXTRATERMS : (sks1)]]