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Proof step 753 from proof hasSharing-put-index-refval (previous step)

(st[r0 -- i ', refval(r2)][r2 -- (sks3 + i ' ' + sks0)]) = refval(r1), st[r -- sks1] = refval(r0), st[r -- sks] = refval(r1), st[r0 -- _type] .type = mkarraytype(ty), st[r2 -- sks3] = refval(r0), 0 < st[r0 -- _length] .intval, fits(r2, sks3 + i ' ' + sks0, st), fits(r, sks1, st), fits(r, sks, st), disjoint(str, r ' + r2 ', st), validref(str, r2, ty, st), i < st[r0 -- _length] .intval, 0 ≤ i, ∀ sks2. sks2 + i ' ' ⊑ sks3 → st[r2 -- sks2] .refval ≠ r0 ⊦ sks1 + i ' ' + sks3 + i ' ' + sks0 = sks, sks1 = [], r2 = jvmref, r0 = jvmref, r = r2, r1 = jvmref, sks = [], junkref(str, r, st), hasSharing(r2, st), hasSharing(r, st), ∃ sks2, sks5. sks3 + i ' ' + sks0 = sks2 + i ' ' + sks5 ∧ st[r2 -- sks2] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks2 → st[r2 -- sks4] .refval ≠ st[r2 -- sks2] .refval), ∃ sks0, sks. sks1 = sks0 + i ' ' + sks ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks0 → st[r -- sks4] .refval ≠ st[r -- sks0] .refval), ∃ sks. ¬ (sks + i ' ' ⊑ sks1 → st[r -- sks] .refval ≠ r0), ∃ sks0, sks1. sks = sks0 + i ' ' + sks1 ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks2. sks2 + i ' ' ⊑ sks0 → st[r -- sks2] .refval ≠ st[r -- sks0] .refval)

User applied rule exists right

And obtained the following new goals

Proof step 752:
(st[r0 -- i ', refval(r2)][r2 -- (sks3 + i ' ' + sks0)]) = refval(r1), st[r -- sks1] = refval(r0), st[r -- sks] = refval(r1), st[r0 -- _type] .type = mkarraytype(ty), st[r2 -- sks3] = refval(r0), 0 < st[r0 -- _length] .intval, fits(r2, sks3 + i ' ' + sks0, st), fits(r, sks1, st), fits(r, sks, st), disjoint(str, r ' + r2 ', st), validref(str, r2, ty, st), i < st[r0 -- _length] .intval, 0 ≤ i, ∀ sks2. sks2 + i ' ' ⊑ sks3 → st[r2 -- sks2] .refval ≠ r0 ⊦ sks1 + i ' ' + sks3 + i ' ' + sks0 = sks, sks1 = [], r2 = jvmref, r0 = jvmref, r = r2, r1 = jvmref, sks = [], junkref(str, r, st), hasSharing(r2, st), hasSharing(r, st), sks3 + i ' ' + sks0 = sks3 + i ' ' + sks0 ∧ st[r2 -- sks3] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks3 → st[r2 -- sks4] .refval ≠ st[r2 -- sks3] .refval), ∃ sks0, sks. sks1 = sks0 + i ' ' + sks ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks4. sks4 + i ' ' ⊑ sks0 → st[r -- sks4] .refval ≠ st[r -- sks0] .refval), ∃ sks. ¬ (sks + i ' ' ⊑ sks1 → st[r -- sks] .refval ≠ r0), ∃ sks0, sks1. sks = sks0 + i ' ' + sks1 ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks2. sks2 + i ' ' ⊑ sks0 → st[r -- sks2] .refval ≠ st[r -- sks0] .refval)

Rule argument

[EXRARG : pos: right 11, [extquanttermlist: (sks3, sks0) abort:F , discard:T , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sks3, sks0)]]