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Proof step 478 from proof hasSharing-put-index-refval (previous step)

(st[r0 -- i ', refval(r4)][r -- (sks2 + i ' ' + sks)]) = refval(r1), st[r -- sks1] = refval(r0), (st[r0 -- i ', refval(r4)][r4 -- sks3]) = refval(r1), st[r0 -- _type] .type = mkarraytype(ty), st[r -- sks2] = refval(r0), 0 < st[r0 -- _length] .intval, fits(r, sks2 + i ' ' + sks, st), is_referencevalue(st[r0 -- i ', refval(r4)][r0 -- (i ' ' + sks3)]), 0 ≤ i, i < st[r0 -- _length] .intval, fits(r4, sks3, st[r0 -- i ', refval(r4)]), fits(r, sks1, st), disjoint(str, r ' + r4 ', st), validref(str, r4, ty, st), ∀ sks0. sks0 + i ' ' ⊑ sks2 → st[r -- sks0] .refval ≠ r0, r4 = jvmref → is_reftype(ty) ⊦ sks2 + i ' ' + sks = sks1 + i ' ' + sks3, sks1 = [], r0 = jvmref, r = r4, r1 = jvmref, junkref(str, r, st), hasSharing(r4, st), hasSharing(r, st), ∃ sks0, sks5. sks2 + i ' ' + sks = sks0 + i ' ' + sks5 ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks0 → st[r -- sks4] .refval ≠ st[r -- sks0] .refval), ∃ sks, sks0. sks1 = sks + i ' ' + sks0 ∧ st[r -- sks] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks → st[r -- sks4] .refval ≠ st[r -- sks] .refval), ∃ sks. ¬ (sks + i ' ' ⊑ sks1 → st[r -- sks] .refval ≠ r0)

User applied rule exists right

And obtained the following new goals

Proof step 477:
(st[r0 -- i ', refval(r4)][r -- (sks2 + i ' ' + sks)]) = refval(r1), st[r -- sks1] = refval(r0), (st[r0 -- i ', refval(r4)][r4 -- sks3]) = refval(r1), st[r0 -- _type] .type = mkarraytype(ty), st[r -- sks2] = refval(r0), 0 < st[r0 -- _length] .intval, fits(r, sks2 + i ' ' + sks, st), is_referencevalue(st[r0 -- i ', refval(r4)][r0 -- (i ' ' + sks3)]), 0 ≤ i, i < st[r0 -- _length] .intval, fits(r4, sks3, st[r0 -- i ', refval(r4)]), fits(r, sks1, st), disjoint(str, r ' + r4 ', st), validref(str, r4, ty, st), ∀ sks0. sks0 + i ' ' ⊑ sks2 → st[r -- sks0] .refval ≠ r0, r4 = jvmref → is_reftype(ty) ⊦ sks2 + i ' ' + sks = sks1 + i ' ' + sks3, sks1 = [], r0 = jvmref, r = r4, r1 = jvmref, junkref(str, r, st), hasSharing(r4, st), hasSharing(r, st), sks2 + i ' ' + sks = sks2 + i ' ' + sks ∧ st[r -- sks2] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks2 → st[r -- sks4] .refval ≠ st[r -- sks2] .refval), ∃ sks, sks0. sks1 = sks + i ' ' + sks0 ∧ st[r -- sks] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks → st[r -- sks4] .refval ≠ st[r -- sks] .refval), ∃ sks. ¬ (sks + i ' ' ⊑ sks1 → st[r -- sks] .refval ≠ r0)

Rule argument

[EXRARG : pos: right 9, [extquanttermlist: (sks2, sks) abort:F , discard:T , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sks2, sks)]]