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Proof step 5 from proof delete-shift (previous step)

a ∈ s, a0 ∈ s, ∀ a0. a0 ≠ a ∧ a0 ∈ s ↔ a0 ∈ s0 ⊦ (a1 ∈ s ↔ a1 ∈ s0 ++ a) ∧ ¬ a ∈ s0

User applied rule all left

And obtained the following new goals

Proof step 6:
a1 ≠ a ∧ a1 ∈ s ↔ a1 ∈ s0, a0 ∈ s, a ∈ s, ∀ a0. a0 ≠ a ∧ a0 ∈ s ↔ a0 ∈ s0 ⊦ (a1 ∈ s ↔ a1 ∈ s0 ++ a) ∧ ¬ a ∈ s0

Rule argument

pos: left 3 terms: a1 abort: F discard: F computed: T

Additional information

Used terms: a1