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Proof step 7 from proof gfts-all (previous step)

∀ sk0. sk0 = sk ∨ sk0 ∈ sks0 → goodfieldandtype(r, sk0, st), (∀ sk. sk ∈ sks0 → goodfieldandtype(r, sk, st)) → goodfieldsandtypes(r, sks0, st), ∀ r, st. (∀ sk. sk ∈ sks0 → goodfieldandtype(r, sk, st)) → goodfieldsandtypes(r, sks0, st) ⊦ goodfieldsandtypes(r, sks0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
sk0 ∈ sks0, ∀ r, st. (∀ sk. sk ∈ sks0 → goodfieldandtype(r, sk, st)) → goodfieldsandtypes(r, sks0, st), ∀ sk0. sk0 = sk ∨ sk0 ∈ sks0 → goodfieldandtype(r, sk0, st) ⊦ goodfieldsandtypes(r, sks0, st), goodfieldandtype(r, sk0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]