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Proof step 13 from proof in-add (previous step)

∀ st, ty, j, r, vals0. # vals0 < # vals1 + 1 → (true → (r -- j ' ∈ addarray(r, ty, vals0, st) ↔ ¬ j < 0 ∧ ¬ n→i(# vals0) ≤ j ∨ r -- j ' ∈ st)) ⊦ r -- j ' ∈ (addarray(r, ty, vals1, st)[r -- n→i(# vals1) ', val]) ↔ r -- j ' ∈ st ∨ ¬ j < 0 ∧ ¬ n→i(# vals1) < j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st
Proof step 12:
∀ st, ty, j, r, vals0. # vals0 < # vals1 + 1 → (true → (r -- j ' ∈ addarray(r, ty, vals0, st) ↔ ¬ j < 0 ∧ ¬ n→i(# vals0) ≤ j ∨ r -- j ' ∈ st)) ⊦ r -- j ' = r -- n→i(# vals1) ' ∨ r -- j ' ∈ addarray(r, ty, vals1, st) ↔ r -- j ' ∈ st ∨ ¬ j < 0 ∧ ¬ n→i(# vals1) < j

Rule argument

[speclemmaarg: ,rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st ,[val, rk0, st, rk] → [val, r -- n→i(# vals1) ', addarray(r, ty, vals1, st), r -- j '] ,F ,store ,javastore ,In-insert ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val, r -- n→i(# vals1) ', addarray(r, ty, vals1, st), r -- j ')]]