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Proof step 8 from proof gfts-all (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 7:
∀ 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)

Rule argument

[EXRARG : pos: left 2, [extquanttermlist: (r, st)abort:F, discard:F, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, st)]]