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Proof step 7 from proof validfieldinits-one (previous step)

fv = sk × val ⊦ validfieldinits(str, fv ', st) ↔ ¬ is_indexkey(sk) ∧ type∃(str, sk .fs .type) ∧ validval(str, val, sk .fs .type, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ validfieldinits(str, (sk × val) ', st) ↔ ¬ is_indexkey(sk) ∧ type∃(str, sk .fs .type) ∧ validval(str, val, sk .fs .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]