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Proof step 14 from proof compatval-put (previous step)

compat(str, refval(r), ty, st) ⊦ ty0 = void_type, (ty = void_type ∨ is_reftype(ty)) ∧ ( r ≠ jvmref → r -- _type ∈ st ∧ is_typevalue(st[r -- _type]) ∧ is_reftype(st[r -- _type] .type) ∧ ( has_classtype(st[r -- _type] .type) → class∃(str, st[r -- _type] .type .jtelemtype .class) ∨ interface∃(str, st[r -- _type] .type .jtelemtype .class) ∨ is_predefined_class(st[r -- _type] .type .jtelemtype .class)) ∧ (is_classtype(st[r -- _type] .type) → class+∃(str, st[r -- _type] .type .class)) ∧ is_subtype(str, st[r -- _type] .type, ty))

User applied rule case distinction

And obtained the following new goals

Proof step 5:
compat(str, refval(r), ty, st) ⊦ ty = void_type ∨ is_reftype(ty), ty0 = void_type
Proof step 13:
compat(str, refval(r), ty, st) ⊦ r ≠ jvmref → r -- _type ∈ st ∧ is_typevalue(st[r -- _type]) ∧ is_reftype(st[r -- _type] .type) ∧ ( has_classtype(st[r -- _type] .type) → class∃(str, st[r -- _type] .type .jtelemtype .class) ∨ interface∃(str, st[r -- _type] .type .jtelemtype .class) ∨ is_predefined_class(st[r -- _type] .type .jtelemtype .class)) ∧ (is_classtype(st[r -- _type] .type) → class+∃(str, st[r -- _type] .type .class)) ∧ is_subtype(str, st[r -- _type] .type, ty), ty0 = void_type

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]