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Proof step 11 from proof putar-inother (previous step)

∀ i, j, r0, r, st. i < j ∨ n→i(# vals0) + j ≤ i → (r0 -- i ' ∈ putArray(vals0, r, j, st) ↔ r0 -- i ' ∈ st) ⊦ i < j ∨ n→i(# (val + vals0)) + j ≤ i → (r0 -- i ' ∈ putArray(val + vals0, r, j, st) ↔ r0 -- i ' ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ putArray(val ' + vals, r, i, st) = (putArray(vals, r, i + 1, st)[r -- i ', val])
Proof step 10:
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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + 1 ≤ j ↔ i < j], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ n→i(m + n) = n→i(m) + n→i(n)], [CSIMPSEQ : ⊦ # val ' = 0 +1], [CSIMPSEQ : ⊦ # (vals + vals0) = # vals + # vals0], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ val + vals = val ' + vals])]]