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Proof step 7 from proof getintarray-in-02 (previous step)

i ∈ getintarray(r, i0, i1, st) ∨ i = st[r -- (i0 + i1) '] .intval, 0 ≤ i1, ∀ st, i0, r, i, j0. i→n(j0) < i→n(i1 + 1) → (true → (i ∈ getintarray(r, i0, j0, st) ↔ (∃ j1. i = st[r -- (i0 + j1) '] .intval ∧ ¬ j1 < 0 ∧ ¬ j0 ≤ j1))) ⊦ i ∈ getintarray(r, i0, i1, st), ∃ j. i = st[r -- (i0 + j) '] .intval ∧ 0 ≤ j ∧ j < i1, ∃ j. i = st[r -- (i0 + j) '] .intval ∧ ¬ j < 0 ∧ ¬ i1 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 ≤ i1, ∀ st, i0, r, i, j0. i→n(j0) < i→n(i1 + 1) → (true → (i ∈ getintarray(r, i0, j0, st) ↔ (∃ j1. i = st[r -- (i0 + j1) '] .intval ∧ ¬ j1 < 0 ∧ ¬ j0 ≤ j1))) ⊦ st[r -- (i0 + i1) '] .intval ∈ getintarray(r, i0, i1, st), ∃ j. st[r -- (i0 + i1) '] .intval = st[r -- (i0 + j) '] .intval ∧ ¬ j < 0 ∧ ¬ i1 ≤ j, ∃ j. st[r -- (i0 + i1) '] .intval = st[r -- (i0 + j) '] .intval ∧ ¬ j < 0 ∧ ¬ i1 < j

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i])]]