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Proof step 11 from proof ao-put (previous step)

∀ r, r0, class1, st, sk, val. r = r0 ∨ (addobj(r, class1, fis0, st)[r0 -- sk, val]) = addobj(r, class1, fis0, st[r0 -- sk, val]) ⊦ r = r0, (addobj(r, class1, fis0, st)[r -- fv .1, fv .2][r0 -- sk, val]) = (addobj(r, class1, fis0, st[r0 -- sk, val])[r -- fv .1, fv .2])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
fv = sk0 × val0, ∀ r, r0, class1, st, sk, val. r = r0 ∨ (addobj(r, class1, fis0, st)[r0 -- sk, val]) = addobj(r, class1, fis0, st[r0 -- sk, val]) ⊦ r = r0, (addobj(r, class1, fis0, st)[r -- sk0, val0][r0 -- sk, val]) = (addobj(r, class1, fis0, st[r0 -- sk, val])[r -- sk0, val0])

Rule argument

[speclemmaarg: ,sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val ,[fv, sk, val] → [fv, sk0, val0] ,T ,pair ,fieldinit ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val])], [EXTRATERMS : (fv, sk0, val0)]]