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Proof step 7 from proof in-init-02 (previous step)

∀ r, sk, r0, j, st. r -- sk ∈ addarray(r0, vals0, j, st) ∧ r ∈ st → r = r0 ∨ r -- sk ∈ st, r -- sk ∈ addarray(r0, val + vals0, j, st), r ∈ st ⊦ r = r0, r -- sk ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ addarray(r, val ' + vals, i, st) = (addarray(r, vals, i + 1, st)[r -- i ', val])
Proof step 5:
⊦ r ∈ st → r ∈ addarray(r0, vals, j, st)
Proof step 6:
r -- sk ∈ addarray(r0, vals0, j + 1, st), r ∈ st, ∀ r, sk, r0, j, st. r -- sk ∈ addarray(r0, vals0, j, st) ∧ r ∈ st → r = r0 ∨ r -- sk ∈ st ⊦ r = r0, r -- sk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st], [CSIMPSEQ : ⊦ r ≠ r0 → ( r -- sk ∈ st[r0 -- sk0, val] ↔ r -- sk ∈ st)], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ val + vals = val ' + vals])]]