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Proof step 19 from proof subset-size (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ st[a, d] ⊆ st0 ↔ st -- a ⊆ st0 -- a ∧ a ∈ st0 ∧ st0[a] = d
Proof step 21:
st0[a] = d, a ∈ st0, # st0 ≤ # st1, st1 -- a ⊆ st0 -- a, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦ st1[a, d] = ∅, a ∈ 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