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Proof step 10 from proof putar-inother (previous step)

n→i(# vals0) + j < i ∨ i < j, ∀ i0, j0, r1, r2, st0. ¬ j0 ≤ i0 ∨ ¬ i0 < n→i(# vals0) + j0 → (r1 -- i0 ' ∈ putArray(vals0, r2, j0, st0) ↔ r1 -- i0 ' ∈ st0) ⊦ r0 -- i ' ∈ (putArray(vals0, r, j + 1, st)[r -- j ', val]) ↔ r0 -- i ' ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ r ≠ r0 → (r -- sk ∈ st[r0 -- sk0, val] ↔ r -- sk ∈ st)
Proof step 8:
¬ r0 ≠ r, n→i(# vals0) + j < i ∨ i < j, ∀ i0, j0, r1, r2, st0. ¬ j0 ≤ i0 ∨ ¬ i0 < n→i(# vals0) + j0 → (r1 -- i0 ' ∈ putArray(vals0, r2, j0, st0) ↔ r1 -- i0 ' ∈ st0) ⊦ r0 -- i ' ∈ (putArray(vals0, r, j + 1, st)[r -- j ', val]) ↔ r0 -- i ' ∈ st
Proof step 9:
⊦ r ≠ r0 → (r0 -- sk ∈ putArray(vals, r, j, st) ↔ r0 -- sk ∈ st)

Rule argument

[speclemmaarg: ,r ≠ r0 → (r -- sk ∈ st[r0 -- sk0, val] ↔ r -- sk ∈ st) ,[val, sk0, r0, st, sk, r] → [val, j ', r, putArray(vals0, r, j + 1, st), i ', r0] ,F ,javastore , ,in-01 ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val, j ', r, putArray(vals0, r, j + 1, st), i ', r0)]]