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Proof step 15 from proof del-exists (previous step)

st0 = st[a, st0[a]], a ∈ st0, ∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]) ⊦ a ∈ st, ∃ st1. ¬ a ∈ st1 ∧ st0[a0, d] = st1[a, st0[a0, d][a]]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ a ≠ b → st[b, d][a] = st[a]
Proof step 17:
¬ a ≠ a0, st0 = st[a, st0[a]], a ∈ st0, ∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]) ⊦ a ∈ st, ∃ st1. ¬ a ∈ st1 ∧ st0[a0, d] = st1[a, st0[a0, d][a]]
Proof step 34:
a ≠ a0, st0 = st[a, st0[a]], a ∈ st0, ∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]) ⊦ a ∈ st, ∃ st1. ¬ a ∈ st1 ∧ st0[a0, d] = st1[a, st0[a]]

Rule argument

spec: inst: name: At-other seq: ⊦ a ≠ b → st[b, d][a] = st[a] subst: [a, d, b, st] → [a, d, a0, st0] flags: rewrite, current

Additional information

Used terms: a, d, a0, st0