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Proof step 13 from proof getarray-union (previous step)

true, ∀ st0, st, i, r, j0. i→n(j0) < i→n(j) → (true → getarray(r, i, j0, st ∪ st0) = getarray(r, i, j0, st0) ∨ r ∈ st) ⊦ getarray(r, i, j, st ∪ st0) = getarray(r, i, j, st0) ∨ r ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ st0, st, i, r, j0. i→n(j0) < i→n(j) → (true → getarray(r, i, j0, st ∪ st0) = getarray(r, i, j0, st0) ∨ r ∈ st) ⊦ getarray(r, i, j, st ∪ st0) = getarray(r, i, j, st0), r ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]