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Proof step 57 from proof validfieldinits-def-01 (previous step)

∀ str, st. validfieldinits(str, fis0, st) ↔ validfieldinits(fis0, st) ∧ ¬ junkrefs(str, fis0 .vals, st) ∧ (∀ sk. sk ∈ fis0 → type∃(str, sk .fs .type)) ⊦ validfieldinits(str, fv ' + fis0, st) ↔ validfieldinits(fv ' + fis0, st) ∧ ¬ junkrefs(str, fv .2 ' + fis0 .vals, st) ∧ (∀ sk. sk ∈ fv ' + fis0 → type∃(str, sk .fs .type))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 56:
fv = sk0 × val, ∀ str, st. validfieldinits(str, fis0, st) ↔ validfieldinits(fis0, st) ∧ ¬ junkrefs(str, fis0 .vals, st) ∧ (∀ sk. sk ∈ fis0 → type∃(str, sk .fs .type)) ⊦ validfieldinits(str, fv ' + fis0, st) ↔ validfieldinits(fv ' + fis0, st) ∧ ¬ junkrefs(str, val ' + fis0 .vals, st) ∧ (∀ sk. sk ∈ fv ' + fis0 → type∃(str, sk .fs .type))

Rule argument

[speclemmaarg: ,sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val ,[fv, sk, val] → [fv, sk0, val] ,T ,pair ,fieldinit ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val])], [EXTRATERMS : (fv, sk0, val)]]