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Proof step 6 from proof validmode-put-nonnull (conclusion)

⊦ ¬ _mode ∈ rkvs ∧ st[_mode] = noval → (validmode(str, st[rkvs]) ↔ validmode(str, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ st[_mode] = noval → (validmode(str, st) ↔ _mode ∈ st)
Proof step 5:
st[_mode] = noval ⊦ validmode(str, st[rkvs]) ↔ _mode ∈ st, _mode ∈ rkvs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]