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Proof step 3 from proof mode-noval-subset (conclusion)

⊦ st ⊆ st0 ∧ _mode ∈ st ∧ st0[_mode] ≠ noval → st[_mode] ≠ noval

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
st[_mode] = noval, _mode ∈ st, st ⊆ st0 ⊦ st0[_mode] = noval

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]