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Proof step 2 from proof eq-shift-put (previous step)

a ∈ st ⊦ st[a, d] = st0 ↔ st = st0[a, st[a]] ∧ a ∈ st0 ∧ st0[a] = d

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 ⊦ (∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0) ∧ st[a, d][a0] = st0[a0]) ↔ st = st0[a, st[a]] ∧ a ∈ st0 ∧ st0[a] = d

Rule argument

spec: store inst: name: Extension seq: ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]) subst: [st2, st1] → [st0, st[a, d]] flags: rewrite, current

Additional information

Used terms: st0, st[a, d]