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Proof step 8 from proof putar-inother (previous step)

¬ r0 ≠ r, 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) ⊦ r0 -- i ' ∈ (putArray(vals0, r, j + 1, st)[r -- j ', val]) ↔ r0 -- i ' ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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 -- j ', val]) ↔ r -- i ' ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]