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Proof step 10 from proof instore-02 (previous step)

r = r0 ⊦ ∃ sk1. r = r0 ∧ sk1 = sk ∨ r -- sk1 ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ ∃ sk0. r0 -- sk0 ∈ st ∨ sk0 = sk

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]