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Proof step 20 from proof hasSharing-put-index-refval (previous step)

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

User applied rule exists right

And closed the goal!

Rule argument

[EXRARG : pos: right 10, [extquanttermlist: (sks0)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 : (sks0)]]