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Proof step 4 from proof putar-inref (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

[emptyarg]

Additional information

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