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Proof step 9 from proof putar-rew (previous step)

r0 -- sk ≠ r -- j ', n→i(# vals0) + j < sk .index ∨ r0 ≠ r ∨ is_fskey(sk) ∨ sk .index < j, ∀ r1, r2, sk0, j0, st0, val1. ¬ is_indexkey(sk0) ∨ r1 ≠ r2 ∨ ¬ j0 ≤ sk0 .index ∨ ¬ sk0 .index < n→i(# vals0) + j0 → (putArray(vals0, r2, j0, st0)[r1 -- sk0, val1]) = putArray(vals0, r2, j0, st0[r1 -- sk0, val1]) ⊦ (putArray(vals0, r, j + 1, st)[r0 -- sk, val0][r -- j ', val]) = (putArray(vals0, r, j + 1, st[r0 -- sk, val0])[r -- j ', val])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n→i(# vals0) + j < sk .index ∨ r0 ≠ r ∨ is_fskey(sk) ∨ sk .index < j, ∀ r1, r2, sk0, j0, st0, val1. ¬ is_indexkey(sk0) ∨ r1 ≠ r2 ∨ ¬ j0 ≤ sk0 .index ∨ ¬ sk0 .index < n→i(# vals0) + j0 → (putArray(vals0, r2, j0, st0)[r1 -- sk0, val1]) = putArray(vals0, r2, j0, st0[r1 -- sk0, val1]), r0 ≠ r ∨ sk ≠ j ' ⊦ (putArray(vals0, r, j + 1, st)[r0 -- sk, val0][r -- j ', val]) = (putArray(vals0, r, j + 1, st[r0 -- sk, val0])[r -- j ', val])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)])]]