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Proof step 6 from proof subset-put (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 7:
a ∈ st[a, d] → a ∈ st0 ∧ d = st0[a], a0 ∈ st[a, d] → a0 ∈ st0 ∧ st[a, d][a0] = st0[a0], ∀ a0. a0 ∈ st[a, d] → a0 ∈ st0 ∧ st[a, d][a0] = st0[a0] ⊦ (a0 ∈ st -- a → a0 ∈ st0 ∧ (st -- a)[a0] = st0[a0]) ∧ a ∈ st0 ∧ st0[a] = d

Rule argument

pos: left 2 terms: a0 abort: F discard: F computed: T

Additional information

Used terms: a0