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Proof step 20 from proof ac-lookup-03 (previous step)

is_fskey(fv .1), ∀ class1, st. (addclass(class1, fis0, st)[jvmref -- mkfs(class1, void_type, `initstate' .field) ']) = initval(done) ∨ mkfs(class1, void_type, `initstate' .field) ' ∈ fis0 ⊦ (addclass(class1, fv ' + fis0, st)[jvmref -- mkfs(class1, void_type, `initstate' .field) ']) = initval(done), mkfs(class1, void_type, `initstate' .field) ' ∈ fv ' + fis0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 19:
fv = sk × val, is_fskey(sk), ∀ class1, st. (addclass(class1, fis0, st)[jvmref -- mkfs(class1, void_type, `initstate' .field) ']) = initval(done) ∨ mkfs(class1, void_type, `initstate' .field) ' ∈ fis0 ⊦ (addclass(class1, fv ' + fis0, st)[jvmref -- mkfs(class1, void_type, `initstate' .field) ']) = initval(done), mkfs(class1, void_type, `initstate' .field) ' ∈ fv ' + 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)]]