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Proof step 20 from proof getintarray-in-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
∀ st, i0, r, i, j0. i→n(j0) < i→n(j) → (true → (i ∈ getintarray(r, i0, j0, st) ↔ (∃ j1. i = st[r -- (i0 + j1) '] .intval ∧ ¬ j1 < 0 ∧ ¬ j0 ≤ j1))) ⊦ i ∈ getintarray(r, i0, j, st) ↔ (∃ j0. i = st[r -- (i0 + j0) '] .intval ∧ ¬ j0 < 0 ∧ ¬ j ≤ j0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]