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Proof step 9 from proof subset-size (previous step)

# st0 < # st1, a ∈ st1, st1[a, d] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦ st1[a, d] = ∅, st1[a] = d, st1 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ st[a, d] ⊆ st0 ↔ st -- a ⊆ st0 -- a ∧ a ∈ st0 ∧ st0[a] = d
Proof step 11:
st0[a] = d, # st0 < # st1, a ∈ st0, a ∈ st1, st1 -- a ⊆ st0 -- a, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦ st1[a] = d, st1[a, d] = ∅, st1 = ∅

Rule argument

spec: <toplevel> name: subset-put-02 seq: ⊦ st[a, d] ⊆ st0 ↔ st -- a ⊆ st0 -- a ∧ a ∈ st0 ∧ st0[a] = d subst: [st0, d, a, st] → [st0, d, a, st1] flags: <all paths>

Additional information

Used terms: st0, d, a, st1