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Proof step 51 from proof putarray-addarray-one (previous step)

is_fskey(sk), r0 -- sk ∈ addarray(r0, ty, n→i(# vals) + 1, st) ∨ ¬ is_fskey(sk) ∧ ¬ sk .index ≤ 0 ∧ ¬ n→i(# vals) < sk .index ↔ r0 -- sk ∈ addarray(r0, ty, refval(jvmref) ' + vals, st), is_reftype(ty) ⊦ (addarray(r0, ty, n→i(# vals) + 1, st)[r0 -- sk]) = (addarray(r0, ty, refval(jvmref) ' + vals, st)[r0 -- sk])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 50:
r0 -- sk ∈ addarray(r0, ty, n→i(# vals) + 1, st) ↔ r0 -- sk ∈ addarray(r0, ty, refval(jvmref) ' + vals, st), is_fskey(sk), is_reftype(ty) ⊦ (addarray(r0, ty, n→i(# vals) + 1, st)[r0 -- sk]) = (addarray(r0, ty, refval(jvmref) ' + vals, st)[r0 -- sk])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]