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Proof step 3 from proof add-in-06 (previous step)

⊦ sk0 ∈ (sk × val) ' + fis0 ∧ (∀ sk, r, class1, st. sk ∈ fis0 → r -- sk ∈ addobj(r, class1, fis0, st)) → r -- sk0 ∈ addobj(r, class1, (sk × val) ' + fis0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ addobj(r, class1, (sk × val) ' + fis, st) = (addobj(r, class1, fis, st)[r -- sk, val])
Proof step 2:
sk0 ∈ fis0, ∀ sk, r, class1, st. sk ∈ fis0 → r -- sk ∈ addobj(r, class1, fis0, st) ⊦ sk0 = sk, r -- sk0 ∈ addobj(r, class1, fis0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)], [CSIMPSEQ : sk ≠ sk0 ⊦ sk ∈ (sk0 × val) ' + fis ↔ sk ∈ fis], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st])]]