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Proof step 6 from proof gfts-put (previous step)

r ≠ r0, ∀ r, r0, st, sk0, val. r = r0 ∨ (goodfieldsandtypes(r, sks0, st[r0 -- sk0, val]) ↔ goodfieldsandtypes(r, sks0, st)) ⊦ r = r0, goodfieldandtype(r, sk, st) ∧ goodfieldsandtypes(r, sks0, st[r0 -- sk0, val]) ↔ goodfieldandtype(r, sk, st) ∧ goodfieldsandtypes(r, sks0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ r, r0, st, sk0, val. r = r0 ∨ (goodfieldsandtypes(r, sks0, st[r0 -- sk0, val]) ↔ goodfieldsandtypes(r, sks0, st)) ⊦ r = r0, goodfieldandtype(r, sk, st) ∧ goodfieldsandtypes(r, sks0, st[r0 -- sk0, val]) ↔ goodfieldandtype(r, sk, st) ∧ goodfieldsandtypes(r, sks0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]