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Proof step 7 from proof update-union-switch (previous step)

∀ r, st, st0. update(r, fis0, st ∪ st0) = st ∪ update(r, fis0, st0) ∨ r ∈ st .refs ⊦ (update(r, fis0, st ∪ st0)[r -- sk, val]) = st ∪ (update(r, fis0, st0)[r -- sk, val]), r ∈ st .refs

User applied rule all left

And obtained the following new goals

Proof step 6:
update(r, fis0, st ∪ st0) = st ∪ update(r, fis0, st0) ∨ r ∈ st .refs, ∀ r, st, st0. update(r, fis0, st ∪ st0) = st ∪ update(r, fis0, st0) ∨ r ∈ st .refs ⊦ (update(r, fis0, st ∪ st0)[r -- sk, val]) = st ∪ (update(r, fis0, st0)[r -- sk, val]), r ∈ st .refs

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (r, st, st0) abort:F , discard:F , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, st, st0)]]