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Proof step 5 from proof mapex-new (previous step)

P(mk(rs, f)) ↔ ((r =>_ * P))(mk(rs, f) ++ r) ⊦ P(mk(rs, f)) ↔ ((r =>_ * P))(mk(rs ++ r, f)), r = null, r ∈ rs

User applied rule insert equation

And obtained the following new goals

Proof step 6:
⊦ ((r =>_ * P))(mk(rs, f) ++ r) ↔ ((r =>_ * P))(mk(rs ++ r, f)), r = null, r ∈ rs

Rule argument

pos: left 1 flags: discard

Additional information