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Proof step 9 from proof Del-at (previous step)

∀ b, a. (st0 -- b)[a] = st0[a] ∨ a = b ⊦ (st0[a0, d] -- b)[a] = st0[a], a = a0, a = b

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ a ≠ b → st[b, d] -- a = (st -- a)[b, d]
Proof step 11:
b = a0, ∀ b, a. (st0 -- b)[a] = st0[a] ∨ a = b ⊦ (st0[a0, d] -- b)[a] = st0[a], a = b, a = a0
Proof step 13:
∀ b, a. (st0 -- b)[a] = st0[a] ∨ a = b ⊦ (st0 -- b)[a0, d][a] = st0[a], a = b, a = a0, b = a0

Rule argument

spec: <toplevel> name: Del-rec-other seq: ⊦ a ≠ b → st[b, d] -- a = (st -- a)[b, d] subst: [a, d, b, st] → [b, d, a0, st0] flags: <all paths>

Additional information

Used terms: b, d, a0, st0