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Proof step 11 from proof update-in-03 (previous step)

∀ r, r0, st. r ∈ update(r0, fis0, st) ↔ r ∈ st ∨ r = r0 ∧ fis0 ≠ [] ⊦ r ∈ (update(r0, fis0, st)[r0 -- fv .1, fv .2]) ↔ r ∈ st ∨ r = r0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
fv = sk × val, ∀ r, r0, st. r ∈ update(r0, fis0, st) ↔ r ∈ st ∨ r = r0 ∧ fis0 ≠ [] ⊦ r ∈ (update(r0, fis0, st)[r0 -- sk, val]) ↔ r ∈ st ∨ r = r0

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)]]