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Proof step 8 from proof inj-put (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 9:
(a0 ∈ st[a, d] ↔ a0 ∈ st0[a, d]) ∧ st[a, d][a0] = st0[a, d][a0], ∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0[a, d]) ∧ st[a, d][a0] = st0[a, d][a0] ⊦ a ∈ st0, a ∈ st, (a0 ∈ st ↔ a0 ∈ st0) ∧ st[a0] = st0[a0]

Rule argument

pos: left 1 terms: a0 abort: F discard: F computed: F

Additional information

Used terms: a0