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Proof step 4 from proof in-1init-01 (previous step)

∀ r, r0, j, st. r = r0 ∨ (r ∈ addarray(r0, vals0, j, st) ↔ r ∈ st) ⊦ r = r0, r ∈ addarray(r0, val + vals0, j, st) ↔ r ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ addarray(r, val ' + vals, i, st) = (addarray(r, vals, i + 1, st)[r -- i ', val])
Proof step 3:
∀ r, r0, j, st. r = r0 ∨ (r ∈ addarray(r0, vals0, j, st) ↔ r ∈ st) ⊦ r = r0, r ∈ addarray(r0, vals0, j + 1, st) ↔ r ∈ st

Rule argument

[emptyarg]

Additional information

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