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Proof step 6 from proof putar-inother (previous step)

i ' ≠ j ', n→i(# vals0) + j < i ∨ i < j, ∀ i0, j0, r1, r2, st0. ¬ j0 ≤ i0 ∨ ¬ i0 < n→i(# vals0) + j0 → (r1 -- i0 ' ∈ putArray(vals0, r2, j0, st0) ↔ r1 -- i0 ' ∈ st0) ⊦ r -- i ' ∈ putArray(vals0, r, j + 1, st) ↔ r -- i ' ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n→i(# vals0) + j < i ∨ i < j, ∀ i0, j0, r1, r2, st0. ¬ j0 ≤ i0 ∨ ¬ i0 < n→i(# vals0) + j0 → (r1 -- i0 ' ∈ putArray(vals0, r2, j0, st0) ↔ r1 -- i0 ' ∈ st0) ⊦ r -- i ' ∈ putArray(vals0, r, j + 1, st) ↔ r -- i ' ∈ st, i = j

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i ' = i0 ' ↔ i = i0])]]