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Proof step 23 from proof ac-in-03 (previous step)

is_fskey(fv .1), jvmref -- _type ∈ addclass(class1, fis0, st), jvmref ∈ addclass(class1, fis0, st), ∀ class, st0. jvmref -- _type ∈ addclass(class, fis0, st0) ∧ jvmref ∈ addclass(class, fis0, st0) → jvmref -- _type ∈ st0 ⊦ fv .1 .fs .class = class1, jvmref -- _type ∈ st, _type ∈ fv ', _type ∈ fis0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 22:
fv = sk × val, is_fskey(sk), jvmref -- _type ∈ addclass(class1, fis0, st), jvmref ∈ addclass(class1, fis0, st), ∀ class, st0. jvmref -- _type ∈ addclass(class, fis0, st0) ∧ jvmref ∈ addclass(class, fis0, st0) → jvmref -- _type ∈ st0 ⊦ sk .fs .class = class1, jvmref -- _type ∈ st, _type ∈ fv ', _type ∈ fis0

Rule argument

[speclemmaarg: ,sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val ,[fv, sk, val] → [fv, sk, val] ,T ,pair ,fieldinit ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val])], [EXTRATERMS : (fv, sk, val)]]