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Proof step 2 from proof Del-in (previous step)

∀ a, b. a ∈ st0 -- b ↔ a ≠ b ∧ a ∈ st0 ⊦ a ∈ st0[a0, d] -- b ↔ a ≠ b ∧ a ∈ st0[a0, d]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ a ≠ b → st[b, d] -- a = (st -- a)[b, d]
Proof step 4:
b = a0, ∀ a, b. a ∈ st0 -- b ↔ a ≠ b ∧ a ∈ st0 ⊦ a ∈ st0[a0, d] -- b ↔ a ≠ b ∧ a ∈ st0[a0, d]
Proof step 6:
∀ a, b. a ∈ st0 -- b ↔ a ≠ b ∧ a ∈ st0 ⊦ a ∈ (st0 -- b)[a0, d] ↔ a ≠ b ∧ a ∈ st0[a0, d], 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