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Proof step 24 from proof putAr-lastrec (previous step)

(r -- i ') .ref ≠ r ∨ is_fskey((r -- i ') .key) ∨ (r -- i ') .key .index < i + 1 ∨ n→i(# vals0) + i + 1 ≤ (r -- i ') .key .index, putArray(vals0 + val ', r, i + 1, st[r -- i ', val0]) = (putArray(vals0, r, i + 1, st)[r -- i ', val0][r -- ((i + 1) + n→i(# vals0)) ', val]), ∀ val, r, i, st. putArray(vals0 + val ', r, i, st) = (putArray(vals0, r, i, st)[r -- (i + n→i(# vals0)) ', val]) ⊦ (putArray(vals0 + val ', r, i + 1, st)[r -- i ', val0]) = (putArray(vals0, r, i + 1, st)[r -- ((1 + n→i(# vals0)) + i) ', val][r -- i ', val0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
putArray(vals0 + val ', r, i + 1, st[r -- i ', val0]) = (putArray(vals0, r, i + 1, st)[r -- i ', val0][r -- ((i + 1) + n→i(# vals0)) ', val]), ∀ val, r, i, st. putArray(vals0 + val ', r, i, st) = (putArray(vals0, r, i, st)[r -- (i + n→i(# vals0)) ', val]) ⊦ (putArray(vals0 + val ', r, i + 1, st)[r -- i ', val0]) = (putArray(vals0, r, i + 1, st)[r -- ((1 + n→i(# vals0)) + i) ', val][r -- i ', val0])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i ≤ i], [CSIMPSEQ : ⊦ i < j + 1 ↔ i ≤ j], [CSIMPSEQ : ⊦ i ' .index = i], [CSIMPSEQ : ⊦ ¬ is_fskey(i ')], [CSIMPSEQ : ⊦ (r -- sk) .key = sk], [CSIMPSEQ : ⊦ (r -- sk) .ref = r])]]