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Proof step 12 from proof update-ex-nodups (previous step)

∃ fis1. ¬ dups(fis1 .keys) ∧ update(r, fis0, st) = update(r, fis1, st) ∧ fis1 ⊆ fis0 ⊦ ∃ fis. ¬ dups(fis .keys) ∧ (update(r, fis0, st)[r -- sk, val]) = update(r, fis, st) ∧ fis ⊆ (sk × val) ' + fis0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
update(r, fis0, st) = update(r, fis, st), fis ⊆ fis0 ⊦ dups(fis .keys), ∃ fis. ¬ dups(fis .keys) ∧ (update(r, fis0, st)[r -- sk, val]) = update(r, fis, st) ∧ fis ⊆ (sk × val) ' + fis0

Rule argument

[emptyarg]

Additional information

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