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Proof step 21 from proof add-dupput (previous step)

¬ r ≠ r0, sk ∈ fis ⊦ (r0 -- sk0 ∈ addobj(r, class1, fis + (sk × val) ' + fis0, st) ↔ r0 -- sk0 ∈ addobj(r, class1, fis + fis0, st)) ∧ (addobj(r, class1, fis + (sk × val) ' + fis0, st)[r0 -- sk0]) = (addobj(r, class1, fis + fis0, st)[r0 -- sk0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ r -- sk ∈ addobj(r, class, fis, st) ↔ ¬ ¬ (sk ∈ fis ∨ sk = _type ∨ r -- sk ∈ st)
Proof step 20:
sk ∈ fis ⊦ (addobj(r0, class1, fis + (sk × val) ' + fis0, st)[r0 -- sk0]) = (addobj(r0, class1, fis + fis0, st)[r0 -- sk0])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ sk ∈ (sk0 × val) ' ↔ sk = sk0], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPSEQ : ⊦ sk ∈ fis + fis0 ↔ sk ∈ fis ∨ sk ∈ fis0])]]