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Proof step 6 from proof add-nput-02 (previous step)

r -- sk ≠ r -- sk0, ∀ r, class1, sk, val, st. addobj(r, class1, fis0 + (sk × val) ', st) = (addobj(r, class1, fis0, st)[r -- sk, val]) ∨ sk ∈ fis0 ⊦ (addobj(r, class1, fis0 + (sk × val) ', st)[r -- sk0, val0]) = (addobj(r, class1, fis0, st)[r -- sk, val][r -- sk0, val0]), sk = sk0, sk ∈ fis0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ r, class1, sk, val, st. addobj(r, class1, fis0 + (sk × val) ', st) = (addobj(r, class1, fis0, st)[r -- sk, val]) ∨ sk ∈ fis0 ⊦ (addobj(r, class1, fis0 + (sk × val) ', st)[r -- sk0, val0]) = (addobj(r, class1, fis0, st)[r -- sk, val][r -- sk0, val0]), sk = sk0, sk ∈ fis0

Rule argument

[emptyarg]

Additional information

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