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Proof step 9 from proof subset-put-01 (previous step)

a ∈ st → a ∈ st0[a, st[a]] ∧ st[a] = d, ∀ a0. a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0] ⊦ (a1 ∈ st → a1 ∈ st0 ∧ st[a1] = st0[a1]) ∧ ¬ a ∈ st, a ∈ st ∧ st[a] = d ∧ (a ∈ st -- a → a ∈ st0 -- a ∧ (st -- a)[a] = (st0 -- a)[a])

User applied rule all left

And obtained the following new goals

Proof step 10:
a ∈ st → a ∈ st0[a, st[a]] ∧ st[a] = d, a1 ∈ st → a1 ∈ st0[a, d] ∧ st[a1] = st0[a, d][a1], ∀ a0. a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0] ⊦ a ∈ st ∧ st[a] = d ∧ (a ∈ st -- a → a ∈ st0 -- a ∧ (st -- a)[a] = (st0 -- a)[a]), (a1 ∈ st → a1 ∈ st0 ∧ st[a1] = st0[a1]) ∧ ¬ a ∈ st

Rule argument

pos: left 2 terms: a1 abort: F discard: F computed: T

Additional information

Used terms: a1