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Proof step 8 from proof addobj-at-this (previous step)

⊦ val .refval -- sk0 = r -- sk, (st[r -- _type, typeval(mkclasstype(class))][r -- sk, val][r -- (sk ' + sk0 ')]) = (st[r -- _type, typeval(mkclasstype(class))][r -- sk, val][val .refval -- sk0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
val .refval ≠ r ∨ sk0 ≠ sk ⊦ (st[r -- _type, typeval(mkclasstype(class))][r -- sk, val][r -- (sk ' + sk0 ')]) = (st[r -- _type, typeval(mkclasstype(class))][r -- sk, val][val .refval -- sk0])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)])]]