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Proof step 92 from proof putarray-addarray-one (previous step)

is_reftype(ty) ⊦ putArray(vals, r, 1, addarray(r, ty, n→i(# vals) + 1, st)) = addarray(r, ty, refval(jvmref) ' + vals, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk])
Proof step 91:
is_reftype(ty) ⊦ ∀ rk. (rk ∈ putArray(vals, r, 1, addarray(r, ty, n→i(# vals) + 1, st)) ↔ rk ∈ addarray(r, ty, refval(jvmref) ' + vals, st)) ∧ (putArray(vals, r, 1, addarray(r, ty, n→i(# vals) + 1, st))[rk]) = (addarray(r, ty, refval(jvmref) ' + vals, st)[rk])

Rule argument

[speclemmaarg: ,st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk]) ,[st2, st1] → [addarray(r, ty, refval(jvmref) ' + vals, st), putArray(vals, r, 1, addarray(r, ty, n→i(# vals) + 1, st))] ,F ,store ,javastore ,Extension ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addarray(r, ty, refval(jvmref) ' + vals, st), putArray (vals, r, 1, addarray(r, ty, n→i(# vals) + 1, st)))]]