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Proof step 10 from proof gfts-all (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
¬ goodfieldandtype(r, sk, st), ∀ r, st. (∀ sk. sk ∈ sks0 → goodfieldandtype(r, sk, st)) → goodfieldsandtypes(r, sks0, st), ∀ sk0. sk0 = sk ∨ sk0 ∈ sks0 → goodfieldandtype(r, sk0, st) ⊦
Proof step 9:
¬ goodfieldsandtypes(r, sks0, st), ∀ r, st. (∀ sk. sk ∈ sks0 → goodfieldandtype(r, sk, st)) → goodfieldsandtypes(r, sks0, st), ∀ sk0. sk0 = sk ∨ sk0 ∈ sks0 → goodfieldandtype(r, sk0, st) ⊦

Rule argument

[CASEDARG : pos: left 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]