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Proof step 7 from proof is_obj-subset (previous step)

st ⊆ st1, class+∃(str, st[r -- _type] .type .class), is_obj(r, st) ⊦ class+∃(str, st1[r -- _type] .type .class) ∧ ¬ is_abstract(st1[r -- _type] .type .class) ∧ is_obj(r, st1), is_abstract(st[r -- _type] .type .class)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ st ⊆ st1 ∧ is_obj(r, st) → is_obj(r, st1)
Proof step 6:
st ⊆ st1, class+∃(str, st[r -- _type] .type .class), is_obj(r, st) ⊦ is_abstract(st[r -- _type] .type .class), class+∃(str, st1[r -- _type] .type .class) ∧ ¬ is_abstract(st1[r -- _type] .type .class)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]