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Proof step 2 from proof butmin-put-01 (previous step)

a = st[a, d0].min ⊦ st[a, d] -- st[a, d].min = st, a ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a])
Proof step 4:
a = st[a, d0].min ⊦ ∀ a0. (a0 ∈ st[a, d] -- st[a, d].min ↔ a0 ∈ st) ∧ (st[a, d] -- st[a, d].min)[a0] = st[a0], a ∈ st

Rule argument

spec: store inst: name: Extension seq: ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]) subst: [st2, st1] → [st, st[a, d] -- st[a, d].min] flags: rewrite

Additional information

Used terms: st, st[a, d] -- st[a, d].min