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Proof step 7 from proof in-1dim-07 (previous step)

i > 0, ∀ st, val, fs, r0, i0. i→n(i0) < i→n(i) → (true → (r0 -- fs ' ∈ addarray(r0, val, i0, st) ↔ r0 -- fs ' ∈ st)) ⊦ r0 -- fs ' ∈ (addarray(r0, val, i - 1, st)[r0 -- (i - 1) ', val]) ↔ r0 -- fs ' ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < i, ∀ st, val, fs, r0, i0. i→n(i0) < i→n(i) → (true → (r0 -- fs ' ∈ addarray(r0, val, i0, st) ↔ r0 -- fs ' ∈ st)) ⊦ r0 -- fs ' ∈ addarray(r0, val, i - 1, st) ↔ r0 -- fs ' ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPSEQ : ⊦ i > j ↔ j < i], [CSIMPSEQ : ⊦ r -- fs ' ∈ st[r0 -- i ', val0] ↔ r -- fs ' ∈ st])]]