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Proof step 4 from proof javavalues-in (previous step)

∀ val. val ∈ mkjavavalues(refs0) ↔ is_referencevalue(val) ∧ val .refval ∈ refs0 ⊦ val = refval(r) ∨ val ∈ mkjavavalues(refs0) ↔ is_referencevalue(val) ∧ (val .refval = r ∨ val .refval ∈ refs0)

User applied rule all left

And obtained the following new goals

Proof step 3:
val ∈ mkjavavalues(refs0) ↔ is_referencevalue(val) ∧ val .refval ∈ refs0, ∀ val. val ∈ mkjavavalues(refs0) ↔ is_referencevalue(val) ∧ val .refval ∈ refs0 ⊦ val = refval(r) ∨ val ∈ mkjavavalues(refs0) ↔ is_referencevalue(val) ∧ (val .refval = r ∨ val .refval ∈ refs0)

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (val)abort:F, discard:F, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val)]]