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Proof step 14 from proof in-1dim-other-index-01 (previous step)

r -- i ' ∈ addarray(r, val, j, st), r ∈ addarray(r, val, j, st), i < 0 ∨ j ≤ i, ∀ st, val, r, i, j0. i→n(j0) < i→n(j) → (true → (i < 0 ∨ j0 ≤ i → (r -- i ' ∈ addarray(r, val, j0, st) ↔ r -- i ' ∈ st))) ⊦ r -- i ' ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ i ≤ 0 → addarray(r, val, i, st) = st
Proof step 13:
¬ j ≤ 0, r -- i ' ∈ addarray(r, val, j, st), r ∈ addarray(r, val, j, st), i < 0 ∨ j ≤ i, ∀ st, val, r, i, j0. i→n(j0) < i→n(j) → (true → (i < 0 ∨ j0 ≤ i → (r -- i ' ∈ addarray(r, val, j0, st) ↔ r -- i ' ∈ st))) ⊦ r -- i ' ∈ st

Rule argument

[crewrite-arg: addarray-base, F, [st, i, val, r] → [st, j, val, r]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, j, val, r)]]