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Proof step 11 from proof getintarray-at (previous step)

¬ n→i(n) < j, 0 < j, n < i→n(j) ⊦ (getarray(r, i, j, st)[n]) .intval = st[r -- (i + n→i(n)) '] .intval, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
j ≤ n→i(n), n < i→n(j), 0 < j ⊦ j = 0, (getarray(r, i, j, st)[n]) .intval = st[r -- (i + n→i(n)) '] .intval, n = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 < n ↔ n ≠ 0], [CSIMPSEQ : 0 ≤ i ⊦ ((* i)) < n→i(n) ↔ i→n(i) < n], [CSIMPFORWARD : ⊦ i ≤ j ∧ 0 < i → 0 < j], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i])]]