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Proof step 3 from proof leq-append-both (previous step)

⊦ (∃ σ. (⌜x⌝ + σ1) + σ = ⌜x⌝ + σ2) ↔ σ1 ≤ σ2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
σ1 ≤ σ2 ⊦ ∃ σ. σ1 + σ = σ2

Rule argument

Additional information

Used simplifier rules: append-cancel-l: ⊦ ⌜x⌝ + σ0 = ⌜x⌝ + σ1 ↔ σ0 = σ1;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%leq-append-right: ⊦ σ1 ≤ σ1 + σ2;