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Proof step 28 from proof putar-getar (previous step)

∀ j, r, i, st. j = n→i(# vals0) → getarray(r, i, j, putArray(vals0, r, i, st)) = vals0 ⊦ j = n→i(# (val + vals0)) → getarray(r, i, j, putArray(val + vals0, r, i, st)) = val + vals0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ putArray(val ' + vals, r, i, st) = (putArray(vals, r, i + 1, st)[r -- i ', val])
Proof step 27:
∀ r0, i0, st0. getarray(r0, i0, n→i(# vals0), putArray(vals0, r0, i0, st0)) = vals0 ⊦ getarray(r, i, 1 + n→i(# vals0), putArray(vals0, r, i + 1, st)[r -- i ', val]) = val ' + vals0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ n→i(m + n) = n→i(m) + n→i(n)], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ # val ' = 0 +1], [CSIMPSEQ : ⊦ # (vals + vals0) = # vals + # vals0], [CSIMPSEQ : ⊦ val + vals = val ' + vals])]]