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Proof step 4 from proof add-in-03 (previous step)

⊦ r -- sk0 ∈ st ∧ r ∈ st ∧ (∀ r, sk, st, r0, class1. r -- sk ∈ st ∧ r ∈ st → r -- sk ∈ addobj(r0, class1, fis0, st)) → r -- sk0 ∈ addobj(r0, class1, (sk × val) ' + fis0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ addobj(r, class1, (sk × val) ' + fis, st) = (addobj(r, class1, fis, st)[r -- sk, val])
Proof step 3:
r -- sk0 ∈ st, r ∈ st, ∀ r, sk, st, r0, class1. r -- sk ∈ st ∧ r ∈ st → r -- sk ∈ addobj(r0, class1, fis0, st) ⊦ r -- sk0 ∈ addobj(r0, class1, fis0, st), r = r0 ∧ sk0 = sk

Rule argument

[emptyarg]

Additional information

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